Why does the charge conjugation of the spinor transform as a spinor? I have come across (in QFT Nutshell, A. Zee) how the charge conjugation of the spinor, $\psi_c \equiv \gamma^2 \psi^*$, transform (where $\gamma^2=\sigma^2\otimes i\tau^2$ is the component of the gamma matrices). 
Under a Lorentz transformation, the spinor transforms as
$$\psi \rightarrow e^{-\frac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu}}\psi$$
where 
$$\sigma^{\mu\nu} = \frac{i}{2}[\gamma^\mu,\gamma^\nu]$$
and $\omega_{\mu\nu}$ is antisymmetric
Complex conjugating, we have
$$\psi^* \rightarrow e^{+\frac{i}{4}\omega_{\mu\nu}(\sigma^{\mu\nu})^*}\psi^* = e^{-\frac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu}}\psi^*$$
Hence,
$$\psi_c\rightarrow\gamma^2 e^{-\frac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu}}\psi^* = e^{-\frac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu}}\psi_c$$
It was not obvious to me in the last equation at all. I think that in order to make the last equation correct, we must have $\gamma^2$ commute with the exponential. Or equivalently,
$$\omega_{\mu\nu}[\sigma^{\mu\nu},\gamma^\lambda] = 0$$
with $\lambda=2$.
But this apparently is not correct to me since
$$[\sigma^{\mu\nu},\gamma^\lambda]=2i(\gamma^\mu\eta^{\nu\lambda}-\gamma^\nu\eta^{\mu\lambda})$$
Using the fact that $\omega_{\mu\nu}$ is antisymmetric, we approach the result
$$\omega_{\mu\nu}[\sigma^{\mu\nu},\gamma^\lambda] = 4i \gamma^\mu\omega_\mu^{\;\;\lambda}\neq0$$
 A: First, note user Chiral Anomaly's comment that ${\sigma^{\mu\nu}}^* \neq -\sigma^{\mu\nu}$. $\gamma^2$ is purely complex in the Dirac basis, so it's the only one that switches sign under complex conjugation. This means that $(\gamma^\mu)^* = \gamma^\mu(1 - 2\delta^{\mu2})$. But we won't actually need this, since we can talk about it more abstractly.
Let $C$ be the charge conjugation matrix, which is unitary. We first note the definition that $C^{-1} \gamma^{\mu} C = -({\gamma^{\mu}})^*$. This means that $C \sigma^{\mu \nu} = \frac{i}{2} [(\gamma^\mu)^*,(\gamma^\nu)^*] C = - (\sigma^{\mu\nu})^* C$, where the minus sign comes from the need to conjugate $i$.
This means that $C(-\frac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu})^* = (-\frac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu})C$. And thus that $C (e^{-\frac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu}})^* = e^{-\frac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu}} C$, which is the necessary property you want.
See Tong's notes for more details. This is discussed there around EQ(4.85)
