The only possibility is $n=2$ (in which case the arrows form a clockwise or counterclockwise loop.)

Clearly the topmost row cannot contain $\uparrow$; suppose it has $k$ arrows $\downarrow$. By condition (1), this means the second row must have $k$ arrows $\uparrow$, but by condition (2) we also know the second row must have $k$ arrows $\downarrow$. By the same argument, we know that rows 3 through $n-1$ must each have $k$ arrows $\uparrow$ and $k$ arrows $\downarrow$, while the bottommost row must have $k$ arrows $\uparrow$ and $0$ arrows $\downarrow$. In other words, over the whole board there are $(n-1)k$ arrows $\uparrow$ and $(n-1)k$ arrows $\downarrow$.

By the same reasoning, if the leftmost column has $\ell$ arrows $\rightarrow$, then by conditions (1) and (3), we know that over the whole board there are $(n-1)\ell$ arrows $\rightarrow$ and $(n-1)\ell$ arrows $\leftarrow$, so the total number of arrows on the whole board is $2(n-1)(k+\ell)$. But the total number of arrows must equal the number of cells, so we must have
$$2(n-1)(k+\ell) = n^2.$$
However, since $\gcd(n, n-1) = 1$, this is impossible when $n > 2$.