Let $U$ be a unitary matrix for which one wants to compute the SVD decomposition. One option is to decompose $U$ using its real and imaginary parts, i.e. $U_R$ and $U_I$ respectively. Then
\begin{equation} \begin{split} U_R &= V_1CX^\dagger,\\ U_I &= V_2SX^\dagger. \end{split} \end{equation}
Now it must be that $V_2=V_1F$ or $V_1=V_2F$, where $F$ is a diagonal matrix with $\pm1$'s on the diagonal. The question is a little bit weird, but suppose that the singular values which are the diagonal elements of $C$ and $S$ are real. Now it must be that
$$U = V_1(C+iFS)X^\dagger,$$
where $C+iFS$ is diagonal unitary. What is the order in which the eigenvectors have to be arranged such that all the equations are fulfilled?
In fact, I know that $U_R=V_1CX^\dagger$ is fulfilled (using numerics), but it's giving me trouble to check that $U_I=V_2SX^\dagger$ and $F$ is diagonal. In some sense, $V_1$ and $X$ (and up to some degree the singular values in $C$) fix the order of the basis in $V_2$, but there is still some room for arbitrary placement of the singular vectors in $V_2$. With some arragenment, $U_I\neq V_2SX^\dagger$ but $F$ is diagonal, and with other $U_I=V_2SX^\dagger$ but $F$ is non-diagonal. A particular example as an answer will suffice, I just guess it will have to reproduce the above "wrong" orderings.