A particular relation between two matrices I'd like to consider the consequences of the following relation between two symmetric square matrices $A$ and $B$
$$SA^{-1}S^T=B$$
$$S^TB^{-1}S=A$$
There are no inversion conditions on the matrix $S$.
Now, say we have the eigenvalues/vectors of $A$, can we say anything about the eigenvalues/vectors of $B$?
Edit: I might also add the conditions on $S$
$$S^TS=aA^2$$
$$SS^T=bB^2$$
where $a$ and $b$ are real numbers.
 A: When $A$ and $B$ are real symmetric matrices of the same sizes, the conditions $SA^{-1}S^T=B$ and $S^TB^{-1}S=A$ imply that $S$ is invertible. Hence the two conditions are equivalent and they simply mean that $A$ is congruent to $B$. It follows from Sylvester's law of inertia that $A$ and $B$ have the same inertia, but the exact values of their eigenvalues are completely unrelated. E.g. any two positive definite matrices are congruent to each other, but they have different eigenvalues in general.
With also the conditions $S^TS=aA^2$ and $SS^T=bB^2$, a more precise relation between the eigenvalues of $A$ and $B$ can be obtained. Since $S$ is invertible, $aA^2$ and $bB^2$ are positive definite. Hence $a$ and $b$ are positive real numbers. By taking determinants on both sides of $SA^{-1}S^T=B,\,S^TS=aA^2$ and $SS^T=bB^2$, we further see that $ab=1$.
Now, from $S^TS=aA^2$, we obtain
$$
S=\sqrt{a}Q|A| \text{ and } S=\sqrt{b}|B|U^T\tag{1}
$$
for some real orthogonal matrices $Q$ and $U$. Thus
$$
B=SA^{-1}S^T
=(\sqrt{a}Q|A|)A^{-1}(\sqrt{a}Q|A|)^T
=aQAQ^T.\tag{2}
$$
That is, $B$ is orthogonally similar to $aA$.
In addition, $(2)$ implies that $\sqrt{b}|B|=\sqrt{a}Q|A|Q^T$, while $(1)$ implies that $\sqrt{b}|B|=SU=\sqrt{a}Q|A|U$. It follows that $U=Q^T$. Thus, assuming that $A$ and $B$ are nonsingular symmetric matrices, the general solution to the set of equations $S^TB^{-1}S=A,\ S^TS=aA^2$ and $SS^T=bB^2$ is given by $b=\frac1a>0,\,S=\sqrt{a}Q|A|$ and $B=aQAQ^T$, where $Q$ is real orthogonal. Unless at least two of the variables among $A,B$ and $S$ are known, $Q$ remains arbitrary and we cannot determine the relationship between the eigenvectors of $A$ and $B$.
