I'll show, that the function $f(x,y)=x+y$ gives a counterexample.
First I assume that the square is a rotation of the square with corners $(1,0)$, $(0,1)$, $(-1,0)$ and $(0,-1)$. Say we rotate this square by an angle $\varphi$. Then the corners are $(cos(\varphi),sin(\varphi))$, $(-sin(\varphi), cos(\varphi))$, $(-cos(\varphi),-sin(\varphi))$ and $(sin(\varphi), -cos(\varphi))$. This square satisfies the given functional equation.
If you have an arbitrary square in the plain, it is obtained from one of the rotations by linear transformations, i.e. by multiplying with scalars and translation. By linearity of $f$ the functional equation also holds here.