Find positive definite $X$ such that $AXA^T=\alpha X$.

If the equation is solvable, by taking determinants on both sides, we must have $\alpha=\det(A)^2$. If this is the case, we may put $B=\det(A)^{-1}A$ and equation reduces to $BXB^T=X$.

It is solvable if and only if $B$ is similar to a real orthogonal matrix.

Suppose $BXB^T=X$. Then $(X^{-1/2}BX^{1/2})(X^{-1/2}BX^{1/2})^T=I$. Hence $X^{-1/2}BX^{1/2}$ is real orthogonal.

Conversely, suppose $B=PQP^{-1}$ for some real orthogonal matrix $Q$. Let $X=PYP^T$ (i.e. $Y=P^{-1}X(P^{-1})^T$). The equation then further reduces to $QY=YQ$. Hence any positive definite matrix $Y$ that commutes with $Q$ will give rise to a solution $X$. In particular, we can always take $Y=I$ to obtain a solution $X=PP^T$.