Solution: We recall that if a group $G$ acts on a set $X$ then for any point $x \in X$, the orbit and the stabilizer of $x$ are defined as follows: \begin{align*} O_x & = \{g \cdot x \mid g \in G\} \\ S_x & = \{g \in G \mid g \cdot x = x\}. \end{align*}

Let us denote the elements of the group by \[ e, r, r^2, r^3, l_v, l_h, l_d, l_{d'}, \] where $r$ is the rotation by $90^\circ$ (counterclockwise) and $l_h, l_v, l_d \text{ and } l_{d'}$ denote the reflections along the vertical, horizontal, diagonal, and anti-diagonal respectively.

The group action can be written in the following table. \[ \begin{array}{c|c|c|c|c} & (1,1) & (1,-1) & (-1,-1) & (-1,1) \\ \hline e & (1,1) & (1,-1) & (-1,-1) & (-1,1) \\\hline r & (-1,1) & (1,1) & (1,-1) & (-1,-1) \\\hline r^2 & (-1,-1) & (-1,1) & (1,1) & (1,-1) \\\hline r^3 & (1,-1) & (-1,-1) & (-1,1) & (1,1) \\\hline l_h & (1,-1) & (1,1) & (-1,1) & (-1,-1) \\\hline l_v & (-1,1) & (-1,-1) & (1,-1) & (1,1) \\\hline l_d & (1,1) & (-1,1) & (-1,-1) & (1,-1) \\\hline l_{d'} & (-1,-1) & (1,-1) & (1,1) & (-1,1) \\\hline \end{array} \] Now it is easy to find the orbit and stabilizer of each point. \begin{align*} O_{(1,1)} & = \{(1,1), (1,-1), (-1,1), (-1,-1)\} \\ S_{(1,1)} & = \{ e, l_d\} \\ O_{(-1,1)} & = \{(1,1), (1,-1), (-1,1), (-1,-1)\} \\ S_{(-1,1)} & = \{ e, l_{d'} \} \\ O_{(-1,-1)} & = \{(1,1), (1,-1), (-1,1), (-1,-1)\} \\ S_{(-1,-1)} & = \{ e, l_d \} \\ O_{(1,-1)} & = \{(1,1), (1,-1), (-1,1), (-1,-1)\} \\ S_{(1,-1)} & = \{ e, l_{d'} \} \end{align*} Note that for each of the points, the size of the orbit is $4$ and the size of the stabilizer is $2$. According to the orbit-stabilizer theorem for a finite group, the size of the orbit times the size of the stabilizer is equal to the order of the group. Thus, the orbit-stabilizer theorem holds for the action of $D_4$ on the vertices of the square.