A matrix $M$ is pseudo-Hermitian if it satisfies
$$M^\dagger = \eta M \eta^{-1},$$
where $\eta$ is a Hermitian invertible matrix. The spectrum of pseudo-Hermitian matrices is either completely real or appears in complex conjugate pairs.
Now, physicists are usually interested in a subset of such matrices called $\mathcal P \mathcal T$ symmetric matrices. If you consider a parametrized family of $\mathcal P \mathcal T$ symmetric matrices given by $H(k)$ they may show an interesting symmetry-breaking behaviour: At values of $k<k_c$, where $k_c$ is some critical value, the spectrum is purely real and all eigenvectors are also $\mathcal P \mathcal T$ symmetric. At values $k>k_c$ the symmetry is said to be broken - the spectrum becomes complex-valued and the eigenvectors lose the symmetry.
My question is: Is this symmetry-breaking behaviour in the spectrum generic to all pseudo-Hermitian matrices? If so is there a corresponding behaviour in the eigenvectors too just as there is in the $\mathcal P \mathcal T$ symmetric case?
The answer is yes. Since every matrix is similar to its transpose, there exists an invertible matrix $K$ such that \begin{equation} M^T = K M K^{-1} \quad \text{or} \quad M^\dagger = K^* M^* K^{*^{-1}} = K^* \mathcal{T}M (K^*\mathcal{T})^{-1}. \end{equation} Using the equation defining pseudo-Hermiticity, we get \begin{equation} M = \eta^{-1}K^* \mathcal{T}M (\eta^{-1}K^*\mathcal{T})^{-1} := S \mathcal{T}M (S\mathcal{T})^{-1}, \end{equation} where $S = \eta^{-1}K^*$. While $S$ may not be unitary, and thus, may not have the physical interpretations that $\mathcal{P}$ has, the mathematical framework of $\mathcal{PT}$-breaking transfers completely. Further details are in this paper.