How to solve $MXM^{-1}=-X^t$ ?
M can be symmetric or anti-symmetric
Motivation: Given an inner product space with metric matrix M, a "rotation" matrix $R(k)$ that preserve norms can be obtained by matrix exponential $R(k)=e^{kX}$ where X is the generator matrix of the rotation.
When metric is diagonal and has same signals, the rotation is circular but when metric has opposite signals the "rotation" is hyperbolic.
I wanna find the generator for any metric matrix given.
Suppose that $X,M$ are of size $n \times n$.
Note that, if no further constraint is placed on $X$, then this is simply a linear system of equations. We could solve this linear system by "brute force", assuming $M$ is sufficiently small.
Note that the equation can be rewritten as $$ MX = -X^TM \implies MX + X^TM = 0. $$ With (column-major) vectorization, we could rewrite this system as $$ [(I \otimes M) + (M^T \otimes I)K^{(n,n)}]\operatorname{vec}(X) = 0, $$ where $I$ denotes a size-$n$ identity matrix, $\otimes$ denotes a Kronecker product, and $K^{(n,n)}$ denotes a commutation matrix.
For a $2 \times 2$ matrix $M$ with entries $m_{ij}$, the above equation is $$ \left[ \pmatrix{ m_{11} & m_{12} & 0 & 0\\ m_{21} & m_{22} & 0 & 0\\ 0 & 0 & m_{11} & m_{12}\\ 0 & 0 & m_{21} & m_{22}} + \pmatrix{m_{11} & m_{21} &0& 0\\ 0 & 0& m_{11} & m_{21}\\ m_{22} & m_{12} & 0 & 0\\ 0 & 0&m_{22} & m_{12} } \right]\pmatrix{x_{11}\\ x_{21}\\ x_{12}\\ x_{22}} = 0 $$
Edit: flipped minus sign into plus sign
