Suppose that I have a matrix with only $0$ and $1$ cells,
$$H = \begin{bmatrix} 1 & 0 & 1\\ 0 & 0 &1 \\ 1& 1 &1 \\ 0& 0 & 1\\ 1& 0 &0 \\ 0& 1 & 0 \end{bmatrix}$$
and denote each Singular Value Decomposition as
$$H = U\Sigma V^{T}$$
where $U,V$ are unitary matrices and $\Sigma$ stores the singular values.
Is it possible to find another matrix $H^{'}$, of the same dimensions as $H$ again with only $0$ and $1$, that has the same singular value matrix $\Sigma$, i.e. $H^{'} = U^{'}\Sigma V^{'{T}}$???
If we didn't had the restriction that the cells of $H^{'}$ have to be either $0$ or $1$ we could simply find two unitary matrices $Q_{1}$ and $Q_{2}$ and calculate $$H^{'}=Q_{1}HQ_{2},$$
where this matrix will have as SVD, $H^{'}=(Q_{1}U)\Sigma (V^{T}Q_{2})$, so it will have the same singular values.
However, I cannot find such matrices that obey the restriction that the elements of $H^{'}$ have to be $0$ or $1$.
The only case that I found is the following,
$H^{'}= I_{6\times 6} \times H \times I_{3\times 3}$
where $I$ are identity matrices. But is this the only way?? Because in that case $H^{'}$ will be equal to $H$ or with the rows of $H$ switched.