Let P and Q be two orthogonal matrices such that it makes sense to calculate PAQ. Show that A and PAQ have the same singular values.
So far, I've come to the fact that the SVD of an orthogonal matrix is the matrix itself, so the statement would be obvious, but I'm not sure if this assumption is correct.
the singular values of $A$ are the square root of nonzero eigenvalues of $AA^T.$ then the singular values of $PAQ$ are the square roots of the nonzero eigenvalues of $PAQ(PAQ)^T = PAA^TP^T.$ but the matrices $AA^T$ and $PAA^TP^T$ are similar so they have the eigenvalues and that gives you the desired result.