Gamma matrices are defined by the Clifford algebra
$$ \{\gamma^\mu, \gamma^\nu\}= 2g^{\mu\nu}\mathbb I_n \,. $$
So, you see the index $\mu$ in $\gamma^\mu$ runs from $0$ upto $D-1$ where $D$ is the number of spacetime dimensions. It does not mean $\gamma^\mu$ is a vector. The $\mu$ index here only tells you how many gamma matrices are there. The dimensionality of the matrices themselves is $n= 2^{[D/2]}$ where $[\cdot]$ gives you the integer part of a number. For example, in $(1+2)-$dimensions, $D=3$ and hence the Dirac matrices are $2^{[1.5]}= 2$ dimensional, which you recognize are the Pauli matrices. The labels of the entries of the gamma matrices are known as spinor indices. So, in 3 dimensions, for example, the $a,b$ in $\gamma^\mu_{ab}$ would run from $1$ to $2$.
What is a $4$-vector? It is something that transforms like a vector under Lorentz transformations $\Lambda$. Namely, $X^\mu$ is a vector if it transforms like
$$ X^\mu\to {\Lambda^\mu}_\nu X^\nu \,. $$
That's the definition! Just having a $4$-dimensional column vector with Greek indices labelling its entries does not make it a Lorentz vector. It needs to transform the right way.
Okay, so what is a spinor? A spinor is something that transforms like a spinor. Namely, $\psi$ is a spinor if it transforms, under a Lorentz transformation parametrized by $\omega_{\mu\nu}$, like
$$ \psi \to \Lambda_{1/2} \psi\, \qquad (\Rightarrow \overline\psi \to \overline\psi\ \Lambda_{1/2}^{-1}\ ) \,, $$
where $\Lambda_{1/2} = \exp{(-\frac i2 \omega_{\mu\nu} S^{\mu\nu})}$ and $S^{\mu\nu} = \frac i4 [\gamma^\mu, \gamma^\nu]$ generates an $n-$dimensional representation of the Lorentz algebra.
Let's make a remark on why we use something like $\overline \psi =
\psi^\dagger \gamma^0$. Well, because we want to construct bilinear
Lorentz scalars like $\psi^\dagger \psi$, but $\psi^\dagger \psi$ is
not a Lorentz scalar precisely because the matrix $\Lambda_{1/2}$ is
not unitary. Under a Loretz transformation,
$$ \psi^\dagger \to \psi^\dagger \Lambda_{1/2}^\dagger \ne
\psi^\dagger \Lambda_{1/2}^{-1}\,.$$
However, we notice an interesting property of the gamma matrix $\gamma^0$.
$$ \boxed{ \Lambda_{1/2}^\dagger \gamma^0 = \gamma^0
\Lambda_{1/2}^{-1} }$$
This immediately tells us that defining something like $\overline \psi
\equiv \psi^\dagger \gamma^0$ will do the job.
$$ \overline \psi \to (\psi^\dagger \Lambda_{1/2}^\dagger)\gamma^0 =
\psi^\dagger \gamma^0 \Lambda_{1/2}^{-1} = \overline\psi
\Lambda_{1/2}^{-1} $$
Because of this special property of $\gamma^0$, now we have that
$\overline\psi\psi\to \overline\psi\psi$.
You can check that the gamma matrices also satisfy the relation
$$ \Lambda_{1/2}^{-1} \gamma^\mu_{ab} \Lambda_{1/2} = {\Lambda^\mu}_\nu \gamma^\nu_{ab}\,. $$
Understand that this is not a transformation of the gamma matrices under a Lorentz transformation. Gamma matrices are fixed constant matrices that form the basis of an algebra. They do not transform. The above is just a property of the gamma matrices due to them being generators of a particular representation of the Lorentz algebra.
However, this relation allows you to take the $\mu$ index in $\gamma^\mu$ "seriously". Because, due to this you can immediately see that under a Lorentz transformation, the current $J^\mu := \overline\psi \gamma^\mu \psi= \overline\psi^a \gamma^\mu_{ab} \psi^b$ indeed transforms like a vector.
$$ J^\mu \to {\Lambda^\mu}_\nu J^\nu \,.$$
In Euclidean $d=2k$ dimensions, we can take $\gamma^1, \gamma^2,\ldots, \gamma^{2k}$ to be hermitian. We can also arrange the usual $a_n, a^\dagger_n$ construction so that the $\gamma^i$ are symmetric for odd $i$ and antisymmetric for even $i$.
Assuming this to be the case, we define
$$
C_1 = \prod_{i{\rm\, odd}} \gamma^i, \quad C_2= \prod_{i{\rm \,even}} \gamma^i,
$$
and use them to construct ${\mathcal C}$ and ${\mathcal T}$ such that
$$
{\mathcal C}\gamma^i {\mathcal C}^{-1} = - (\gamma^i)^T,
$$
$$
{\mathcal T}\gamma^i {\mathcal T}^{-1} = + (\gamma^i)^T.\nonumber
$$
We find that for
{k=0, mod 4}:
${\mathcal C}=C_1$ symmetric, ${\mathcal T}=C_2$ symmetric. Both commute with $\gamma^{2k+1}$.
{k=1, mod 4}:
${\mathcal C}=C_2$ antisymmetric, ${\mathcal T}=C_1$ symmetric. Both anticommute with $\gamma^{2k+1}$.
{k=2, mod 4}:
${\mathcal C}=C_1$ antisymmetric, ${\mathcal T}=C_2$ antisymmetric. Both commute with $\gamma^{2k+1}$.
{k=3, mod 4}:
${\mathcal C}=C_2$ symmetric, ${\mathcal T}=C_1$ antisymmetric. Both anticommute with $\gamma^{2k+1}$.
Under a change of basis $\gamma^\mu \to A\gamma^\mu A^{-1}$ for the gamma matrices, ${\mathcal C}$ and ${\mathcal T}$ will no longer given by the explicit product expressions $C_1$ and $C_2$, but instead they transform as
$$
{\mathcal C} \to A^T {\mathcal C} A,\quad {\mathcal T} \to A^T {\mathcal T} A.
$$
The symmetry or antisymmetry of ${\mathcal C}$, ${\mathcal T}$ is unchanged, and is is thus a basis-independent property. Another way to think of this is by making use of the transpose of the the defining transformations. We then see that ${\mathcal C}^{-1}{\mathcal C}^T$ and ${\mathcal T}^{-1}{\mathcal T}^T$ commute with all $\gamma^\mu$ and so, by Schur's lemma, both ${\mathcal C}, {\mathcal T}$ are proportional to their transpose. Then, transposing again,
$$
C= \lambda C^T \Rightarrow C= \lambda^2 C
$$
so $\lambda=\pm 1$ in any representation. Since the Euclidean $\gamma^\mu$ are Hermitian, a similar argument shows that ${\mathcal C}^\dagger {\mathcal C}$ is proportional to the identity. As ${\mathcal C}^\dagger {\mathcal C}$ is a positive operator, the factor of proportionality is positive --- so ${\mathcal C}, {\mathcal T}$ can be scaled by a real factor so as to be unitary in all representations.
In odd dimensions we have to ask ourselves whether these objects commute or anticommute with the extra gamma matrix. We can construct (but not easy to display in MathJax) a table that displays the outcome of this examination. It shows whether the matrices exist, and whether they are symmetric (S), or antisymmetric (A). The table repeats mod 8.
When ${\mathcal T}$ is symmetric we can find a basis in which ${\mathcal T}= {\mathbb I}$ and all the Euclidean gamma matrices are real. When ${\mathcal C}$ is symmetric we can find a basis in which all the gamma matrices are pure imaginary.
Charge conjugation: We define the charge conjugate fields by
$$
\psi^c= {\mathcal C}^{-1}{\bar \psi}^T= {\mathcal C}^{-1}(\gamma^0)^T \psi^*,
$$
$$
\bar \psi^c = -\psi^T {\mathcal C}.\nonumber
$$
The transformation of $\bar \psi$ (which in Euclidean space is independent of $\psi$) is chosen to be consistent with the Minkowski version. To obtain this Minkowski signature version we first note that we can use the same ${\mathcal T}$ and ${\mathcal C}$ matrices as in Euclidean space --- the inclusion of factors of $i$ in the transposing $\gamma^\mu$ does not change the formula. Consider, for example, the mostly minus Minkowski metric $(+,-,-,\ldots)$ in which $\gamma^0$ is Hermitian and obeys $(\gamma^0)^2=1$. Then
$$
\psi^c= {\mathcal C}^{-1} (\gamma^0)^T \psi^*\, \Rightarrow (\psi^c)^\dagger = \psi ^T(\gamma^0)^T {\mathcal C},
$$
as ${\mathcal C}$ remains unitary in Minkowski space. Then we have
$$
\overline{(\psi^c)}\equiv (\bar \psi)^c = (\psi^c)^\dagger \gamma^0
$$
$$
= \psi^T (\gamma^0)^T {\mathcal C} \gamma^0 $$
$$
=- \psi^T \,{\mathcal C} \gamma^0 {\mathcal C}^{-1}{\mathcal C} \gamma^0$$
$$
= -\psi^T \,{\mathcal C}.\nonumber
$$
From these results, and with anticommuting $\psi$'s, we find that
$$
\bar \psi^c \gamma^\mu\psi^c =[ - \psi^T {\mathcal C}]\gamma^\mu [{\mathcal C}^{-1} (\gamma^0)^T\psi^*]= \psi^T (\gamma^\mu)^T(\gamma^0)^T \psi^* = - \psi^*\gamma^0 \gamma^\mu \psi =- \bar \psi \gamma^\mu\psi.
$$
Similarly, we have
$$
\bar \psi^c \psi^c =- \psi^T \bar \psi^T = \bar\psi\psi.
$$
The spin-current density transforms as
$$
\bar \psi^c \gamma^0 [\gamma^i,\gamma^j]\psi^c= -\bar\psi \gamma^0 [\gamma^j, \gamma^i]\psi = \bar \psi \gamma^0 [\gamma^i,\gamma^j]\psi.
$$
So, as a result of multiplying six (!) minus signs, the spin is left unchanged.
Using the anticommuting property of the Fermi fields, we now find that, in both Euclidean and Minkowski signatures,
$$
S= \int d^n x\, \bar \psi [\gamma^\mu (\partial_\mu+A)+m]\psi = \int d^n x\, {\bar\psi}^c [\gamma^\mu (\partial_\mu-A^T)+m]\psi^c.
$$
The $-A^T$ are the Lie algebra representation-valued fields in the the conjugate representation to that of $A$, and so $\psi_c$ has the opposite ``charge'' to $\psi$.
Minkowski-signature Marjorana Fermions: We have defined
$$
\psi^c= {\mathcal C}^{-1} (\gamma^0)^T \psi^*
$$
so, with ${\mathcal C}^T=\lambda {\mathcal C}$ we find
$$
(\psi^c)^c= {\mathcal C}^{-1} (\gamma^0)^T ({\mathcal C}^{-1} (\gamma^0)^T\psi^*)^*$$
$$
= {\mathcal C}^{-1}(\gamma^0)^T {\mathcal C}^T\gamma^0 \psi$$
$$
= \lambda \,{\mathcal C}^{-1}(\gamma^0)^T {\mathcal C}\gamma^0 \psi$$
$$
= -\lambda \gamma^0 \gamma^0\psi$$
$$
= -\lambda \psi.$$
We can therefore consistently impose the Majorana condition that $\psi^c=\psi$ only if ${\mathcal C}$ is antisymmetric: i.e. in 2, 3, 4 (mod 8) dimensions. An alternative way to view this is to regard the map
$\psi\to \psi^c$ as an antilinear map $J: {\rm Pin} \to {\rm Pin} $ where ${\rm Pin}$ is the gamma-matrix representation space. If $J^2=id$, this is real structure on the complex ${\rm Pin}$ space.
We can also define another conjugate
$$
\psi^t= {\mathcal T}^{-1}{\bar \psi}^T
$$
$$
\bar \psi^t = \psi^T {\mathcal T}.\nonumber
$$
This reverses the current, again leaves the spin unchanged, but flips the sign of $\bar\psi \psi$. If we demand that $(\psi^t)^t=\psi$, almost identical algebra shows that this defines a real structure only when ${\mathcal T}$ is symmetric hence in d= 8, 9 (mod 8). (The case d=10 (mod 8) is Majorana Weyl.) Some people regard gamma-matrix representations with this real structure also as being Marjorana. I think that these are what Jose} Figueroa-O'Farril calls pseudo-Majorana. Because this``conjugation'' flips $\bar\psi\psi$, these pseudo-Majorana fermions are necessarily massless. It is unfortunate that in the mostly-minus West-cast metric the gamma matrices of a pseudo-Majorana representation can be chosen to be real, while in a Majorana they can only be chosen pure imaginary. It is the other way around in the mostly-plus East-coast metric.
Euclidean Marjorana Fermions: Suppose that we have an eigenfunction of the (skew-adjoint) Dirac operator $D=\gamma^\mu \partial_\mu$ such that
$$
{D}u_n=i\lambda_n u_n,
$$
then
$$
{D}^* u^*=-i\lambda u^*_n.
$$
Assuming that any gauge fields are in real representations, this can be written as
$$
{\mathcal C} {D} {\mathcal C}^{-1} u^*_n = i\lambda_n u^*_n,
$$
or
$$
{D} {\mathcal C}^{-1} u^*_n = i\lambda_n {\mathcal C}^{-1} u^*_n.
$$
Thus $u_n$ and ${\mathcal C}^{-1} u_n^*$ are both eigenfunctions of ${D}$ with the same eigenvalue. They will be linearly independent when ${\mathcal C}$ is antisymmetric --- something that happens in $2,3,4$ (mod 8) Euclidean dimensions. These are precisely the dimensions in which {\it Minkowski space/} Majorana spinors can occur.
This suggests that we take the
Euclidean Majorana-Dirac action to be
$$
S[\psi]= \frac 12 \int d^nx \,\psi^T {\mathcal C}( {D}+m)\psi
$$
and setting
$$
\psi (x) = \sum_n[ \xi_n u_{n}(x)+ \eta_n( {\mathcal C}^{-1}u_{n}^*(x)],
$$
$$
\psi^T(x) = \sum_n [\xi_n u^T_n(x) - \eta_n (u_n^\dagger(x) {\mathcal C}^{-1})],\nonumber
$$
so that
$$
S= \frac 12 \int d^nx \sum_{i,j} ( \xi_i u^T_i - \eta_i u_i^\dagger {\mathcal C}^{-1}) {\mathcal C}(i\lambda_j+m)( \xi_j u_j+ \eta_j {\mathcal C}^{-1}u_j^*)
$$
$$
= \frac 12 \int d^nx \sum_{i, j}\left\{ \xi_i\eta_j (u^T_i u_j^*) + \xi_j \eta_i (u_i^\dagger u_j)\right\}(i\lambda_j+m)
$$
$$
= \sum_i \xi_i\eta_i (i\lambda_i+m).\nonumber
$$
Grassmann integration uses only one copy of the doubly degenerate eigenvalue, and so gives the square-root of the full Dirac determinant.
Best Answer
Quick answer:
We can always define the Dirac adjoint of a spinor to be $\bar \psi := \psi^\dagger \gamma^0$. After that definition is set up, we have to additionally tell how it transforms under a Lorentz transformation and that depends on convention. For example, we could choose $\gamma'^\mu = \Lambda^\mu{}_\nu \gamma^\nu$ (note $\gamma'^0$ is no longer Hermitian) and define $\bar \psi := \psi^\dagger \gamma'^0 $, then we will see that $\bar\psi\psi$ no longer transforms as a scalar.
On the other hand, if we insist that $\bar\psi \psi$ should transform as a scalar, we should find a function of the gamma matrices $F(\gamma'^\mu)$ such that for any Lorentz transformation $\Lambda,\ \color{red}{S(\Lambda)^\dagger FS(\Lambda) \equiv F(\gamma'^\mu)}$. Then, defining $\bar \psi := \psi^\dagger F$ ensures that $\bar\psi\psi$ transforms as a scalar. The question is whether such a function always exists. With your example, you can convince yourself that the ansatz $F= c_\mu \gamma'^\mu$ will work if and only if $c_\mu \Lambda^\mu{}_j=0$ for all spatial indices $j$. This is a system of three equations $(j=1,2,3)$ with four unknowns $(c_\mu)$, and therefore possibly has many solutions. For example, if $\Lambda$ describes a boost in the $x$-direction with speed $\beta$, defining $\bar \psi := \psi^\dagger (\gamma'^0 + \beta \gamma'^1)$ makes $\psi^\dagger\psi$ a scalar. Note that $\gamma'^0 + \beta \gamma'^1 \propto \gamma^0$, so there is no contradiction.
Detailed answer:
A Clifford algebra over a $D$-dimensional spacetime equipped with the metric $g^{\mu\nu}$ is generated by $D$ hypercomplex numbers $\{\gamma^\mu\}, \mu\in \{0,\cdots,D-1\}$ defined by the following algebraic product:
$$ \{\gamma^\mu,\gamma^\nu\} = 2g^{\mu\nu} \mathbb I_n\,. \tag1 $$
The algebra is $2^D$-dimensional, meaning there is a list of $2^D$ linearly independent elements, closed under multiplication, formed by various products of the $D$ hypercomplex numbers. Moreover, there is always a representation of the algebra in terms of real $n \times n$ matrices where $n=2^{[D/2]}$, the operation $[\cdot]$ taking the integer part of the given number. If the dimension $D$ is even, then this representation is the one and only irreducible representation of the algebra (up to equivalences). Furthermore, if the representation is unitary, the $2^D$ basis elements can be chosen to be Hermitian.
Without loss of generality, take the metric to be of mostly negative signature, so that $(\gamma^0)^2 = +1$ and $(\gamma^i)^2=-1$. You already begin to notice why $\gamma^0$ is special when contrasted with the rest of $\gamma$-matrices. It is exactly the same way as time is special when contrasted with space, given that the metric has a Lorentzian signature. With this in mind, for D=4 a basis for $4\times4$ matrices is given by $\{\Gamma_j\}, j\in\{1,\cdots, 16\}$ where
\begin{align*} \Gamma_j \in \{&\mathbb I_4, \gamma^0,i\gamma^1,i\gamma^2,i\gamma^3,\\ &\gamma^0\gamma^1,\ \gamma^0\gamma^2,\ \gamma^0\gamma^3,\quad i\gamma^1\gamma^2,\ i\gamma^2\gamma^3,\ i\gamma^3\gamma^1,\ \\ &i\gamma^0\gamma^1\gamma^2,\quad i\gamma^0\gamma^1\gamma^3,\quad i\gamma^0\gamma^2\gamma^3,\quad i\gamma^1\gamma^2\gamma^3,\\ &i\gamma^0\gamma^1\gamma^2\gamma^3\}\,. \tag2 \\ \end{align*}
The imaginary number $i$ has been inserted in the above array so that for all $j, (\Gamma_j)^2=+1.$ Notice that this list is closed under multiplication. This allows you to write any $4\times4$ matrix $X$ as a sum over these matrices: $X = \sum_{i=1}^{16} x^i \Gamma_i$ where $x_i = \frac14 \text{Tr}(X\Gamma_i)$. Moreover, for all $j\ne1, \text{Tr}(\Gamma_j)= 0$. Using this information, you can prove the following lemma (Cf. Schwartz Ch. 4).
You have correctly identified that $\gamma'^\mu := \Lambda^\mu{}_\nu \gamma^\nu$ obeys the same commutation relation as $\gamma^\mu$, and it is precisely because of this that the above lemma tells us of the existence of a non-singular $S$ so that
$$ S^{-1} \gamma^\mu S = \Lambda^\mu{}_\nu \gamma^\nu\,, \tag3$$
which is required for the Lorentz invariance of the Dirac equation.
At this point, let us make a conventional choice (which @akhmeteli has been pointing out all along). Let us choose $\gamma^0$ such that it is Hermitian and the $\gamma^i$s such that they are anti-Hermitian. In other words, this means that $\gamma^\mu = \gamma^0 (\gamma^\mu)^\dagger \gamma^0$. Notice that we did not need any such assumption in the previous discussion. But this convention will simplify greatly what is about to follow.
Observe that because $\Lambda^\mu{}_\nu$ is real, and because we chose this convention,
\begin{align*} (S^\dagger\gamma^0) \gamma^\mu (S^\dagger \gamma^0)^{-1} = (S^{-1} \gamma^\mu S)^\dagger &\stackrel{(3)}{=} \Lambda^\mu{}_\nu (\gamma^\nu)^\dagger\\ \text{(Multiplying by $\gamma^0$ from both sides)} \Rightarrow (\gamma^0 S^\dagger\gamma^0) \gamma^\mu (\gamma^0 S^\dagger \gamma^0)^{-1} &= \Lambda^\mu{}_\nu \gamma^\nu \stackrel{(3)}{=}S^{-1} \gamma^\mu S\,. \\ \end{align*}
After rearranging we find that,
\begin{align*} &\Rightarrow (S\gamma^0 S^\dagger\gamma^0) \gamma^\mu (S\gamma^0 S^\dagger \gamma^0)^{-1} = \gamma^\mu \\ &\Rightarrow S\gamma^0 S^\dagger\gamma^0 = c\mathbb I_4 \\ &\Rightarrow S^\dagger\gamma^0 = c \gamma^0 S^{-1}\,, \tag4 \\ \end{align*}
where $c$ is some constant which you can convince yourself is real.
Now, if we normalize $S$ such that $\det(S)=1$, then $c^4 = 1$ or $c= \pm 1$. Let us see which situations correspond to $c=+1$ and $c=-1$. Observe the following identity.
\begin{align*} S^\dagger S = (S^\dagger\gamma^0)\gamma^0S &\stackrel{(4)}= c\gamma^0(S^{-1} \gamma^0 S)\\ &\stackrel{(3)}= c\gamma^0 \Lambda^0{}_\nu \gamma^\nu \\ &= c(\Lambda^0{}_0 \mathbb I_4 - \sum_{k=1}^3 \Lambda^0{}_k\gamma^0\gamma^k)\\ \Rightarrow \text{Tr}(S^\dagger S) &= 4c\Lambda^0{}_0 \,. \tag5 \end{align*}
Since $S^\dagger S$ has real eigenvalues (it is Hermitian) and positive-definite (since S is non-singular), its trace must be positive. This means that when $ \Lambda^0{}_0 \le -1, c=-1$ and when $ \Lambda^0{}_0 \ge +1, c=+1$.