Does congruence preserve the reality of matrix eigenvalues?

Question

Let $A \in \mathbb{R}^{n\times n}$ have real eigenvalues. Does \begin{equation} B = X^TAX \end{equation} also have real eigenvalues, for $X \in \mathbb{R}^{n\times n}$ and invertible?

Incidentally the $X$ with which I am working is symmetric and positive definite if either of these would be helpful added structure to adapt the question.

Answer 1

No. Counterexample: $$ A=\pmatrix{1&-3\\ 3&9}, \ B=\pmatrix{1\\ &\frac13}A\pmatrix{1\\ &\frac13}=\pmatrix{1&-1\\ 1&1}. $$ The eigenvalues of $A$ are $5\pm\sqrt{7}$ and the eigenvalues of $B$ are $1\pm i$.