Suppose $\Sigma^{p\times p}$ is positive definite. For $A^{p\times p}$, the solutions to $A^T\Sigma A=I$ can be can be given by what? Suppose $\Sigma$ has eigendecomposition $\Sigma=PLP^T$. Then $P^T\Sigma P=L=L^{1/2}L^{1/2}$. So $L^{-1/2}P^T\Sigma PL^{-1/2}=I$ and a solution for A is $A=PL^{-1/2}$. In fact, multiplying this on the right by any unitary matrix also works. For any unitary matrix U, $U^TL^{-1/2}P^T\Sigma PL^{-1/2} U=I$. So a subset of possible solutions for $A$ are $A=PL^{-1/2}U$ with P being the eigenvectors of A, L being $diag(\lambda)$, and U being any unitary matrix.
Are these the only solutions? If so, prove. If not, find a counterexample such that $A^T\Sigma A=I$ but $PL^{-1/2}U\ne A$ for any choice of unitary matrix $U.$