"Self-congruent" matrices examples

Question

Let's consider just (square) matrices with real entries. Any matrix is congruent to itself, i.e. $$A=B^TAB$$ for $B=I$. But we have also the case of orthogonal matrices $O$, where (e.g.) $I=O^TIO$, or symplectic matrices $S$ where $J=S^TJS$ ($J$ is the standard symplectic matrix). Except these examples, are there other "notable" cases of this relation?

Answer 1

Look at the vectorized form of the problem:

$$ \text{vec}\, A = \text{vec}\,(B^TAB) = (B^T \otimes B^T) \text{vec}\, A $$

Rearranging, we get the homogeneous linear equation:

$$ [I_{n^2} - (B^T \otimes B^T)] \text{vec} A = 0 $$

where $n$ is the number of rows and columns in $A$ and $B$. The only way this can have a solution other than $A = 0$ is if the matrix $[I_{n^2} - (B^T \otimes B^T)]$ has a zero eigenvalue. The eigenvalues of this matrix are

$$ \text{eig}\, [I_{n^2} - (B^T \otimes B^T)] = \{1 - \lambda\mu,\ \lambda,\mu \in \text{eig} B\} $$

In other words, your equation has a solution other than $A = 0$ if and only if the product of any two (not necessarily distinct) eigenvalues of $B$ equals one. I'm not entirely sure what you mean by "notable" cases of the equation, but this is certainly a place to start looking for other interesting examples.

Answer 2

Here $A\in M_n(\mathbb{C})$. We considered the non-linear equation $A=X^TAX$ in the unknown $X\in M_n(\mathbb{C})$.

When $A$ is a generic matrix (for example, randomly choose it), it can be proved that the set of solutions $X$ is an algebraic set of dimension $\left\lfloor n/2\right\rfloor$ (that is, $X$ depends on $\left\lfloor n/2\right\rfloor$ algebraically independent parameters).

The examples you give ($O(n)$ and the symplectic matrices) are very special. They are groups composed by matrices that preserve a non-degenerate bilinear form.