We colour each side of a square with $k \geq 1$ colours. Find a formula for the number of orbits under the action of $D_{4}=\{ e , r,r^{2},r^{3},s,sr,sr^{2},sr^{3} \}$ on the set of colours.
Now as far as I know an orbit of an element $a \in A$ under the action of group $G$ is defined as the set $ G \cdot a = \{ g \cdot a : g \in G\}$. In our case $A$ is the colouring of the square.
My attempt: Now if we number the sides of the square we can represent them as a set $A=\{1,2,3,4\}$.
Now the generators of $G$ are $\langle r \rangle$ and $\langle s \rangle$. No matter the original state of the square $\langle r \rangle $ can only rotate it (by $90$ degrees), and thus the orbit then contains $4$ elements (since $r^{4}=e)$. Now this is also dependent on the amount of colours used (if $k=1$, then the orbit just contains the original square).
As you probably can tell, I don't really grasp the problem yet. Any suggestions are welcome.