Does the following type of "SVD" exist? SVD of $A$ gives $U$, $\Sigma$ and $V$ such that $A = U \Sigma V$.
I am interested in a different problem. Given an $A$ and $\Sigma_1,\ldots,\Sigma_{n-1}$ diagonal matrices, such that we know that
$$A = U_1 \Sigma_1 U_2 \Sigma_2 U_3 \Sigma_3 \ldots U_n$$ for some unitary matrices $U_1,\ldots,U_n$, we need to recover $U_1,\ldots,U_n$.
EDIT: I can choose $\Sigma_i$ freely a priori, if that helps make the recovery of $U$ possible. I also don't mind if the $U$'s are identified up to a multiplication by $-1$ or $1$.

In general this can't be done.  For example, in the $2 \times 2$ case with $V = \pmatrix{0 & 1\cr 1 & 0\cr}$ and $\Sigma_3  = \Sigma_1$,
$$ I \Sigma_1 I \Sigma_2 V \Sigma_1 V = V \Sigma_1 V \Sigma_2 I  \Sigma_1 I $$
In addition, you can always multiply the $U$'s by scalars of absolute value $1$ whose product is $1$.
