I think it is possible. Let $S_n$ be the Symetric group, then, the table with $+1$, determines a permutation $\pi$, and the table with $-1$ determiens a permutation $\sigma$. Since changing lines is equivalnet to do transpositions, and transpositions generate $S_n$, we have that we can acomplish it if there exist permutations $a$, and $b$ such that: $a\pi b=\sigma$ and $a\sigma b=\pi$ but this happens if and only if: $a(\pi\sigma^{-1})^{-1}a^{-1}=(\pi\sigma^{-1})$ and $b^{-1}(\pi^{-1}\sigma)^{-1}b=(\pi^{-1}\sigma)$ But since every permutation is in the same conjugacy class as its inverse (since they have the same cycle lenghts) we can find such $a$ and $b$.

Yes, I believe it's always possible to reach the opposite ($n\times n$) table. If $A$ is our initial matrix, basically, what we want is to find two permutation matrices $P,Q$ such that $PAQ+A=0$. First permute the rows so as to bring all $1$’s on the main diagonal. The $-1$’s lie on the squares $(i,\tau(i))$, where $\tau\in S_n$ is a permutation of $\overline{1,n}$. Since we can use induction on the number of cycles in this permutation, we may as well assume it’s an $n$-cycle $(\sigma(1),\ldots,\sigma(n)),\ \sigma\in S_n$. Let $U$ be the matrix obtained from the identity matrix by permuting its columns according to $\sigma^{-1}$. Then $UAU^t$ ($A$ is not the old matrix $A$, but the one obtained after performing the operations above) will have $1$ on positions $(i,i)$, and $-1$ on positions $(i,i+1)$ (all indices are modulo $n$). This means that we amy as wel assume that our matrix $A$ had this form to begin with. Now let $\tau_1,\tau_2\in S_n$ be the permutations $i\mapsto n-i$ and $i\mapsto n+1-i$ respectively (again, everything is modulo $n$), and let $P,Q$ be the permutation matrices which have $1$ on positions $(i,\tau_1(i))$ (for $P$) and $(\tau_2(j),j)$ (for $Q$). Then $PAQ$ will be $-A$, as desired. Edit: Oops