I wish to solve following second-order matrix equation: $S^HS=Q$, where $\phantom{a}^H$ means complex conjugate, $S,Q$ are square $n\times n$ matrices, and $Q$ must obviously be Hermitian, or else the equation wouldn't be solvable. Furthermore, I wish that the solution be symmetric, i.e. $S^T=S$.
My efforts have led me thus sofar: $Q$ is diagonalizable by a unitary matrix $V$, hence $S^HS=VDV^H$, where $D$ is diagonal. An obvious symmetric solution is $S=V^*D^{1/2}V^H$. There are generally $2^n$ such solutions, since each element of the diagonal matrix $D$ has two square roots (let's assume non-zero eigenvalues for $Q$).
However, if $U$ is an arbitrary unitary matrix, then $US$ is also a solution of $S^HS=Q$. The problem is that $S$ has to be symmetric, which leads to following condition for $U$: $SU^*=US$ or equivalently $S=USU^T$. The problem is that I don't know, if there are any unitary matrices that can satisfy above condition.
My question is: how can one parametrize or describe the complete set of solutions $S$ to the initial problem? If the set of unitary matrices satisfying $SU^*=US$ is in any way parametrizable or describable, this would generate another set of solutions, although I do not know if it would exhaust the possible solutions.
I am thankful for any suggestions!
By Takagi factorisation, if $S$ is complex symmetric, $S=U\Sigma U^T$ for some unitary matrix $U$ and some singular value matrix $\Sigma$. Therefore $S^HS=\bar{U}\Sigma^2U^T$. It follows that $S=U\Sigma U^T$ is a solution to $Q=S^HS$ if and only if $Q=\bar{U}\Sigma^2U^T$ is a unitary diagonalisation. There are not any other solutions.
Edit. Note, however, that there are in general infinitely many solutions: if $S$ is a solution, so is $\omega S$ for any complex number $\omega$ of unit modulus. Put it another way, if $U$ is unitary, so is $\omega^{1/2}U$ for any $\omega$ lying on the unit circle and any square root $\omega^{1/2}$ of it.