Given the SVD $A = U\Sigma V^T$, is $\Sigma$ uniquely determined up to permuting the rows and columns?
My take is that singular value matrix is uniquely determined. Its diagonal elements are square roots of eigenvalues of $A^TA$ and $AA^T$ (both of which have the same unique set of eigenvalues). Thus, $A$ has a unique set of singular values.
Permuting rows and columns of $\Sigma$ we just rearrange the singular values along the diagonal of $\Sigma$. SVD will continue to hold if we rearrange corresponding vectors of $U$ and $V$.
However, such reasoning seems to lack rigour.