Closed form solution for $XAX^{T}=B$

$B^{-1/2}XAX^TB^{-1/2}=I$, so $B^{-1/2}XA^{1/2}=Q$ must be orthogonal. On the other hand, for any orthogonal $Q$, it is simple to verify that $X = B^{1/2}QA^{-1/2}$ solves the equation, so this is a complete parametrization of the solutions.

Here $A^{1/2}$ is the symmetric square root of $A$ (if you prefer you can work with the Cholesky factor and obtain similar results).