Fix a field $k$, and suppose $\gamma$ is an involutory automorphism of $\gamma$ (that is, $\gamma \ne 1$, but $\gamma^2 = 1$).
Call a matrix $A$ $\gamma$-Hermitian if ${(A^\gamma)}^T = A$ (where the "$T$" denotes the transpose).
In case $k = \mathbb{C}$, we know that there is an orthogonal base of eigenvectors spanning $\mathbb{C}^n$, with $A$ an $(n \times n)$-matrix.
Question: is there any general information available about the (possible) eigenvalues of $A$ (for a general field $k$) ?
What about when (modest) assumptions are made about $k$ ?
A property of the complex numbers that is used to establish that eigenvalues of Hermitian matrices are real is that $v^* v = 0$ only if $v = 0$, where $v \in \mathbb C^n$. If this is not the case in field $k$ for the involution $\gamma$ (i.e. there are nonzero vectors such that $(v^\gamma)^T v = 0$), you might have $\gamma$-Hermitian matrices with eigenvalues not invariant under $\gamma$.
For example, take $k = \mathbb Q(\sqrt{2})$ with the involution $\gamma(a+b\sqrt{2}) = a - b \sqrt{2}$. The vector $v = \pmatrix{1\cr 1 + \sqrt{2}}$ satisfies $$ (v^\gamma)^T v = 1^2 + (1 - \sqrt{2})(1+\sqrt{2}) = 0$$ Correspondingly, the $\gamma$-Hermitian matrix $$ A = \pmatrix{-1 & 1\cr 1 & 1\cr}$$ has $v$ as eigenvector for eigenvalue $\sqrt{2}$.
EDIT: However, any eigenvalue $\lambda$ which has an eigenvector $v$ such that $(v^\gamma)^T v \ne 0$ will be invariant under $\gamma$. See the comment below by lonza leggiera.