The answer is all $n$ which are $0$ or $1$ mod $4$. If $n=4k$, simply take the squares with centers at $(1,0),(2,0),\dots,(k,0),(-1,0),\dots,(-k,0)$ and $(0,1),\dots,(0,k),(0,-1),\dots,(0,-k)$. For $n=4k+1$, add the square centered at $(0,0)$. To see why only these work, first note that the axis of symmetry must map the line segments of the squares to other line segments. Since reflection doubles angles, we get that such a symmetry axis must be parallel to $x=0$, $y=0$ or $x=y$, $x=-y$. If we have three of these axes as symmetry axes and they intersect at $P$, then by a suitable reflection, we find that we actually have four axes of symmetry, all passing through $P$. Let $P$ be the origin. Now any point $(a,b)$ except for the origin has orbit $8$ or $4$ under the action of reflection across these lines, so that $n$ is either congruent to $0$ or $1$ mod $4$.