Not in general. First, observe that if we let $M$ be the identity matrix then the identity reduces to $XX^T = I$. In particular, $X \in O(n)$ so $X = X^{-1}$ and thus the map $M \mapsto XMX^T$ is an automorphism of $SO(n)$. But it is also the map $M \mapsto M^T = M^{-1}$ which is an automorphism if and only if a group is Abelian.