= Solution
First conjugate the <Dirac equation>: $(-i\gamma^{\mu*}\partial_\mu-m)\psi^*=0$. A matrix $\Gamma$ taking $\psi^*$ into another solution of the same <Dirac equation> must obey
$$
\gamma^\mu\Gamma=-\Gamma\gamma^{\mu*}.
$$
Because $\gamma^1$ anticommutes with the real $\gamma^0,\gamma^2$ and commutes with the purely imaginary $\gamma^1$, a unitary choice is
$$
\boxed{\Gamma=\gamma^1=i\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad
\psi_C=\gamma^1\psi^*.}
$$
Any constant overall phase is equally valid. To verify invariance of the <action>, rather than only of its equation, keep the <fermion> exchange signs. Define the <charge-conjugation matrix> $C=\Gamma\gamma^0=\sigma_2$. It obeys
$$
C^T=-C,\qquad C^{-1}=C,\qquad
C^{-1}\gamma^\mu C=-(\gamma^\mu)^T,
\qquad
\psi_C=C\bar\psi^T,\quad \bar\psi_C=-\psi^TC^{-1}.
$$
For a classical <Grassmann field>, interchanging two components contributes a minus sign; equivalently use <normal ordering> for quantum <fermionic operators>. Thus
$$
\bar\psi_C\psi_C=-\psi^T\bar\psi^T=\bar\psi\psi,
$$
and
$$
\bar\psi_Ci\gamma^\mu\partial_\mu\psi_C
=i\psi^T(\gamma^\mu)^T(\partial_\mu\bar\psi)^T
=-i(\partial_\mu\bar\psi)\gamma^\mu\psi
=\bar\psi i\gamma^\mu\partial_\mu\psi
-i\partial_\mu(\bar\psi\gamma^\mu\psi).
$$
The last term integrates to a boundary term, vanishing with the usual decay or compatible boundary conditions. Both the <kinetic term> and the <Dirac mass term> therefore leave $S$ invariant under <charge conjugation>. Treating the components as commuting numbers would give the wrong mass-bilinear sign; the <fermion> statistics is essential.
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