Charge conjugation symmetry operation on single-particle Hamiltonian How can I show that given the second-quantized Hamiltonian of a system of non interacting fermions 
$\hat{\mathcal{H}}=\sum_{\alpha, \beta}\hat{\Psi}_{\alpha}^{\dagger}H_{\alpha\beta}\hat{\Psi}_{\beta}$
which is particle-hole symmetric, i.e. $\hat{\mathcal{C}}\hat{\mathcal{H}}\hat{\mathcal{C}}^{-1}=\hat 
{\mathcal{H}}$, with the action of $\hat{\mathcal{C}}$ being defined as
$\hat{\mathcal{C}}\hat{\Psi}_{\alpha}^{\dagger}\hat{\mathcal{C}}^{-1}=\sum_{\beta}\hat{\Psi}_{\beta}(U^{*})_{\beta\alpha}$, 
 $\hat{\mathcal{C}}\hat{\Psi}_{\alpha}\hat{\mathcal{C}}^{-1}=\sum_{\beta}({U^{*}}^{\dagger})_{\alpha\beta}\hat{\Psi}_{\beta}^{\dagger}$,
the single particle Hamiltonian fulfills
$U{H}^{*}U^{\dagger}=-H$?
 A: I'll do it the other way round and show that, given the one-particle Hamiltonan condion, the many-particle Hamiltonian is left invariant.
Let us supose that $U H^* U^{-1}= -H$ and compute 
$$
{\mathcal C}\Psi^\dagger_\alpha H_{\alpha\beta} \Psi_\beta {\mathcal C}^{-1}\\
={\mathcal C}\Psi^\dagger_\alpha {\mathcal C}^{-1}{\mathcal C}H_{\alpha\beta}{\mathcal C}^{-1}{\mathcal C}  \Psi_\beta {\mathcal C}^{-1}\\
= {\mathcal C}\Psi^\dagger_\alpha {\mathcal C}^{-1}H_{\alpha\beta}{\mathcal C} \Psi_\beta {\mathcal C}^{-1}\\
=\Psi_{\rho} U^*_{\rho\alpha} H_{\alpha\beta}  U^{*\dagger}_{\beta\sigma} \Psi^\dagger_\sigma\\
=- \Psi^\dagger_\sigma U^*_{\rho\alpha} H_{\alpha\beta} U^{*\dagger}_{\beta\sigma} \Psi_{\rho}\\
=- \Psi^\dagger_\sigma U_{\sigma\beta} H^T_{\beta \alpha}U^\dagger_{\alpha\rho}\Psi_{\rho}\\
- \Psi^\dagger_\sigma U_{\sigma\beta} H^*_{\beta \alpha}U^\dagger_{\alpha\rho}\Psi_{\rho}\\
= \Psi^\dagger_\sigma H_{\sigma\rho} \Psi_{\rho}.
$$
We have used $U^{*\dagger} = U^T$ and the hemiticity of $H$. So the one-particle transformation on $H$ makes the many particle hamiltonian invariant.
Note that the many-body map ${\mathcal C}$ is a linear map:
$$
{\mathcal C}(\lambda |\psi_1\rangle+\mu |\psi_2\rangle)= \lambda {\mathcal C}|\psi_1\rangle+\mu {\mathcal C}|\psi_2\rangle,
$$
on the Fock space
despite the appearance of "$*$" in the action  on $H$.   This shows up in the step 
$$
{\mathcal C}H_{\alpha\beta}{\mathcal C}^{-1}= H_{\alpha\beta}.
$$
