Show a matrix is anti-symmetric matrix

Question

Let $A\in R^{n\times n}$ be a symmetric matrix, invertible. Let $X\in R^{n\times n}$ be a matrix. I was wondering is $B=A^{-1}XA-X$ always an anti-symmetric matrix? Namely, is $B^T=-B?$ I think the answer is yes, but I don't know how to show it formally.

Answer 1

$$\begin{pmatrix}1&0\\0&2\end{pmatrix}\begin{pmatrix}2&2\\2&2\end{pmatrix}\begin{pmatrix}1&0\\0&\frac12\end{pmatrix}-\begin{pmatrix}2&2\\2&2\end{pmatrix}=\begin{pmatrix}0&-1\\2&0\end{pmatrix}.$$

Answer 2

$$B^T = (A^{-1}XA-X)^T = (A^{-1}XA)^T - X^T = A^TX^T(A^{-1})^T-X^T = A^TX^T(A^T)^{-1}-X^T$$

If $A$ is orthogonal and $X$ is anti-symmetric, then $B$ would also be anti-symmetric.