This Wikipedia page explains that for each of the four main gamma matrices $\gamma^{\mu}$, you can find the covariant matrices $\gamma_{\mu}$ with the equation $\gamma_{\mu} = \eta_{\mu\nu}\gamma^{\mu}$. But that formula doesn't make any sense for $\gamma^5$ because $\eta_{\mu\nu}$ does not have that many indices. So what is $\gamma_5$?
[Physics] the gamma five matrix $\gamma_5$
definitiondirac-equationnotation
Related Solutions
The most general Lorentz transformation that is connected to the identity is given by the conjugation by $\exp(-A)$ where $$ A = \frac 12 \omega_{\mu\nu} \gamma^\mu \gamma^\nu $$ and $\omega_{\mu\nu}$ is an antisymmetric tensor containing $D(D-1)/2$ parameters. The group of all such transformations is isomorphic to $Spin(D-1,1)$. If $\omega$ only contains one component $0\mu$, then it is a boost, and the nonzero numerical value of $\omega$ is the rapidity - the "hyperbolic angle" $\eta$ such that $v/c=\tanh\eta$.
If only one doubly spatial component of $\omega$ is nonzero, then this component $\omega_{\mu\nu} = -\omega_{\nu\mu}$ is obviously the angle itself. Note that the spatial-spatial terms in $A$ are anti-Hermitean, producing unitary transformations; the mixed temporal-spatial terms in $A$ are Hermitean and they don't product unitary transformations on the 4-component space of spinors (but they become unitary if they're promoted to transformations of the full Hilbert space of quantum field theory).
In 4 dimensions, a general antisymmetric matrix $4\times 4$ contains 6 independent parameters and has eigenvalues $\pm i a, \pm i b$, so in 3+1 dimensions, one can always represent a general Lorentz transformation as a rotation around an axis in the 4-dimensional space followed by a boost in the complementary transverse 2-plane. This is the counterpart of the statement that any $SU(2)$ rotations in 3 dimensions is a rotation around a particular axis by an angle.
If you allowed $A$ to contain something else than $\gamma^{\mu\nu}$ matrices which generate the Lorentz group, you could get other groups. Only for a properly chosen subset of allowed values of $\omega$, you would get a closed group from the resulting exponentials (under multiplication). In particular, if you allowed $A$ to be an arbitrary complex combination of any products of gamma matrices, well, then you would allow $A$ to be any complex $4\times 4$ matrix, and its exponentials would produce the full group $GL(4,C)$ - surprising, Carl? ;-) It's not a terribly useful groups in physics because actions are usually not invariant under this "full group", are they?
Also, there are not too many groups in between $Spin(3,1)$ and $GL(4,C)$ - I guess that there's no proper group of $GL(4,C)$ that has a proper $Spin(3,1)$ subgroup. Obviously, there are many subgroups of $Spin(3,1)$ - such as $Spin(3)$, $Spin(1,1)\times Spin(2)$, and others.
How about just testing the two different cases?
I.e. if $\mu\not=0$ then the LHS becomes
\begin{equation} (\gamma^\mu)^\dagger= (\gamma^i)^\dagger= -\gamma^i \tag{see below} \end{equation} while the RHS becomes
\begin{equation} (\gamma^\mu)^\dagger=\gamma^0\gamma^i\gamma^0 = -\gamma^0\gamma^0\gamma^i=-\gamma^i~~~~~~~~ (\text{OK}). \end{equation}
For $\mu=0$, the case is trivial.
EDIT: because of the comment by the OP I will add the following to the answer:
The properties of the gamma matrices can be derived from the properties of the $\vec{\alpha},\beta$-matrices. Some of the properties of the $\vec{\alpha},\beta$ matrices are imposed upon them motivated by physics arguments, such that the Dirac hamiltonian must be hermitian which implies $\vec{\alpha},\beta$ hermitian etc.
Notice that $$ \gamma^\mu := (\beta, \beta\vec{\alpha}).$$
For instance: $$(\gamma^i)^\dagger = (\beta\alpha^i)^\dagger = (\alpha^i)^\dagger\beta^\dagger=\alpha^i\beta=-\beta\alpha^i=-\gamma^i\tag{QED}$$
See page 10 in this PDF for more on the Dirac equation, or see the old book "Quarks and Leptons..." by Halzen and Martin.
See also the wiki page below: http://en.wikipedia.org/wiki/Gamma_matrices#Normalization
Best Answer
The 'five' in $\gamma_5$ is not a Lorentz index, so it doesn't make sense to lower or raise it. It can be defined in different ways, one convention is: $$\gamma_5 = \frac{i}{24}\epsilon_{\mu\nu\rho\sigma}\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma} = \frac{i}{24}\epsilon^{\mu\nu\rho\sigma}\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma_{\sigma}$$, where epsilon is the totally antisymmetric tensor.