• Doesn't condition (4) make all entries of the matrix equal? To get $M_{i,j} = M_{k,\ell}$, act by $k-i$ on the rows and by $\ell-j$ on the columns. – Zach Teitler Feb 6 at 6:04
• @BrendanMcKay Oh, that makes sense. Now it seems to me condition (3) is automatically satisfied: $M_{i,j} = M_{i-j,0}$, so the $i$-th row is a permutation of the $0$-th (first) row; and so is the $j$-th column. So if the first row sums to zero, so do automatically every row and column. – Zach Teitler Feb 6 at 21:02