= Solution
Use the PDF's convention in which $C$ multiplies $\psi^*$, rather than $\bar\psi^T$. Its unitarity converts the given <gamma matrix> identity into
$$
\gamma^\mu C=-C(\gamma^\mu)^*.
$$
Applying it twice gives
$$
M^{\rho\sigma}C
=\frac14[\gamma^\rho,\gamma^\sigma]C
=C\frac14[(\gamma^\rho)^*,(\gamma^\sigma)^*]
=C(M^{\rho\sigma})^*.
$$
For real infinitesimal <Lorentz transformation> parameters,
$$
C\left(I+\frac12\Omega_{\rho\sigma}M^{\rho\sigma}\right)^*
=\left(I+\frac12\Omega_{\rho\sigma}M^{\rho\sigma}\right)C.
$$
Therefore \b[$\psi^c=C\psi^*$ transforms in exactly the same spinor representation as $\psi$]. The coordinate pullback is real and transforms identically as well. This is the <antilinear charge-conjugation intertwiner> relation; exponentiating also gives $CS^*=SC$ for connected transformations.
For real mass, complex conjugation of the free <Dirac equation> gives $(-i\gamma^{\mu*}\partial_\mu-m)\psi^*=0$. Multiply by $C$ and move it through the <gamma matrices>:
$$
0=C(-i\gamma^{\mu*}\partial_\mu-m)\psi^*
=(i\gamma^\mu\partial_\mu-m)C\psi^*.
$$
Thus
$$
\boxed{(i\gamma^\mu\partial_\mu-m)\psi^c=0.}
$$
As an explicit check, in the given <Weyl representation of the gamma matrices> one may take $C=i\gamma^2$. It is unitary and obeys the required conjugation identity. The free-equation conclusion does not assert that a charge-conjugated field retains the same charge in a fixed electromagnetic background.
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