Given matrix $B$ of size $n\times n$ with rank $r>1$.
Denote $S_r$ as $n\times n$ permutation matrix that shifts each row one step down while elevating last row to first.
Denote $S_c$ as $n\times n$ permutation matrix that shifts each column one step right while elevating last column to first.
Form $n^2\times n^2$ block matrix $W$ with $i,j$th block of matrix being $B_{ij}=S_r^{j-1}BS_c^{i-1}$. (Note $j-1$ is for row rotations).
Total number of matrices $B_{ij}$ is $n^2$.
$W$ satisfies row sum = column sum.
Is $\mathsf{rank}(W)\geq rn$?
No. Let $B$ be the matrix of all $1$'s. Then $B_{ij}$ is always the matrix of all $1$'s, and so is $W$. In this case, $r=1$, but $\operatorname{rank}(W) = 1$.