Charge conjugation in arbitrary basis Consider the matrix $C = \gamma^{0}\gamma^{2}$.
It is easy to prove the relations
$$C^{2}=1$$
$$C\gamma^{\mu}C = -(\gamma^{\mu})^{T}$$
in the chiral basis of the gamma matrices.



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*Do the two identities hold in any arbitrary basis of the gamma matrices?

*How is $C$ related to the charge conjugation operator?
 A: Let's instead use the Majorana basis for Gamma matrices, which I'll denote by a tilde. The main thing about this basis is all the gamma matrices are imaginary so the Dirac equation $(i\tilde{\gamma}^{\mu}\partial_\mu-m)\tilde{\psi}=0$ is real, and solutions can be broken up into purely real and imaginary parts. So if $\tilde{\psi}$ satisfies the equation so does $\tilde{\psi}_c\equiv\tilde{\psi}^{*}$. This is what charge conjugation looks like in this basis.
In a different basis of gamma matrices, say the chiral basis, we need to do a unitary transformation $\psi=U\tilde{\psi}$. Then
$$\psi_c=U\tilde{\psi}^*=U(U^\dagger \psi)^*=UU^T \psi^*\equiv\gamma^0C\psi^*,$$
where in the last line we defined the matrix $C$ $$\gamma^0C\equiv UU^T.$$
So above is the formula for charge conjugation in an arbitrary basis, where $C$ is defined in terms of the unitary transformation from the Majorana basis.
The reason we include the factor of $\gamma^0$ is so $C$ satisfies the second identity you mentioned. This follows from $\gamma^0\gamma^\mu\gamma^0=\gamma^{\mu\dagger}$ which is preserved by unitary transformations.
From this we can prove the generalization of the identities you listed
$$C\gamma_\mu C^{-1}=-\gamma_\mu^T$$
$$-CC^*=1$$
What is not necessarily general is $C=C^{-1}$. I presented the same argument here as in the paper arXiv:1006.1718 so you might want to look at that.
