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).
Lemma:
Given two sets of gamma matrices $\{\gamma^\mu\}$ and $\{\gamma'^\mu\}$, correspondingly basis vectors $\{\Gamma_j\}$ and
$\{\Gamma'_j\}$, there exists a non-singular matrix $S$ such that
$$ \boxed{\gamma'^\mu = S\gamma^\mu S^{-1}}\,, \qquad \text{ where }
S=\sum_{i=1}^{16} \Gamma'_i F \Gamma_i\,, $$
and $F$ is so chosen that $S$ is non-singular.
Furthermore, $S$ is
uniquely fixed (up to a numerical factor).
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$.
Conclusion:
Defining $\bar \psi := \psi^\dagger \gamma^0$, we see that under a
Lorentz transformation (taking $\psi \to S\psi$) that does not involve
time reversal ($\Lambda^0{}_0 \ge +1$), $\bar \psi \to +\bar\psi
S^{-1}$, whereas for Lorentz transformations that do reverse time
($\Lambda^0{}_0 \le -1$), $\bar \psi \to -\bar\psi S^{-1}$.
If we had not normalized $S$, we would write $\bar\psi \to c\bar\psi
S^{-1}$. Then this extra factor of $c$ will need taking care of in the
Lagrangian through perhaps a field redefinition.
More generally, if we did not assume the hermiticity properties of the
gamma matrices, things would be much more complicated, but a
definition is a definition. After defining $\bar \psi := \psi^\dagger
\gamma^0$, our task would be to find its transformation rules and then
implement it accordingly in the Lagrangian. Alternatively, if you wish to
preserve the transformation rules, you will have to change the
definition accordingly.
Best Answer
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.
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.
$$ J^\mu \to {\Lambda^\mu}_\nu J^\nu \,.$$