quantum mechanics - Particle hole symmetry in 2nd quantization

In second quantization one the Particle hole trasnformation is defined as

$$\begin{align} \hat{\mathcal{C}} \hat{\psi}_A \hat{\mathcal{C}}^{-1} &= \sum_B U^{*\dagger}_{A,B} \hat{\psi}^{\dagger}_B \\ \hat{\mathcal{C}} \hat{\psi}_A^{\dagger} \hat{\mathcal{C}}^{-1} &= \sum_B \hat{\psi}_B U^{*}_{B,A} \\ \hat{\mathcal{C}} i \hat{\mathcal{C}}^{-1} &= +i \end{align}$$ And if in a 2nd quantized Fermionic Hamiltonian ($\hat{\mathcal{H}}$) Particle Hole symmetry is present then $$\hat{\mathcal{C}} \hat{\mathcal{H}} \hat{\mathcal{C}}^{-1} = \hat{\mathcal{H}}$$ I want to see what this equation means in single particle basis. In single particle basis I can write the 2nd quantized Hamiltonian ($\hat{\mathcal{H}}$) as $$\hat{\mathcal{H}}=\sum_{A,B}\hat{\psi}^\dagger_{A}H_{A,B}\hat{\psi}_B$$ Here the matrix $H$ is the Hamiltonian in single particle basis. Now, with the transformation rules on should get $$U H^{*} U^{\dagger} = - H$$ In the single-particle basis. But what I am getting using the transformation rules is $$U^* H U^{*\dagger} = -H$$ Now I have started to think whether the transformation rules given here are right or not. I wanted to know if the transformation rule or my calculation is wrong.
Source: Topological phases: Classification of topological insulators and superconductors of non-interacting fermions, and beyond Equation 17