You have six constraints, one for each row and column, but the sum of the row constraints and the sum of the column constraints are the same, so at most $5$ of the constraints are linearly independent, so the space is at least $4$-dimensional. The matrix of the first five constraints is
$ \pmatrix{ 1&1&1&0&0&0&0&0&0\\ 0&0&0&1&1&1&0&0&0\\ 0&0&0&0&0&0&1&1&1\\ 1&0&0&1&0&0&1&0&0\\ 0&1&0&0&1&0&0&1&0\\ }\;. $
We can guess three basis vectors by taking the three pairs of indices and placing a $\pmatrix{1&-1\\-1&1}$ matrix in the corresponding entries; that yields
$ \pmatrix{ 1&-1&0&-1&1&0&0&0&0\\ 1&0&-1&0&0&0&-1&0&1\\ 0&0&0&0&1&-1&0&-1&1 } \;. $
Now we can take the generalized cross product of these $8$ vectors, both to check that they're all linearly independent and to find the fourth basis vector that's orthogonal to all of them. According to Wolfram|Alpha, the result is (proportional to)
$ \pmatrix{0&1&-1&-1&0&1&1&-1&0}\;, $
so the matrix for one possible basis is
$ \pmatrix{ 1&-1&0&-1&1&0&0&0&0\\ 1&0&-1&0&0&0&-1&0&1\\ 0&0&0&0&1&-1&0&-1&1\\ 0&1&-1&-1&0&1&1&-1&0 } \;. $