Orbits of rotations under action by D4. If D4 is acting on the subgroup of its rotations, C4, by conjugation, what are the orbits?
I believe that the orbit of each rotation is itself and its own inverse rotation and nothing else. For the identity, it's just the identity.
Is this correct? What's a good way to show/ conclude this? I determined this through drawing out the actions.
 A: $D_n = <r,s | srs = r^{-1}, r^n = s^2 = 1>$
$srs = r^{-1}$, $r_1r_2r_1^{-1} = r_2$ since the rotations form a cyclic group, which is in particular abelian. So the orbits should just be of the form $\{r, r^{-1}\}$ for every rotation $r$.
A: $D_4$ is generated by a rotation $r$ of order $4$, and a reflection $s$ of order $2$. 
As you noted, the identity is in its own one-point orbit.
Observe that $\langle r \rangle$ has index $2$, hence is normal in $D_4$. As $\langle r^2 \rangle$ is characteristic in $\langle r \rangle$ (all subgroups of a finite cyclic group are characteristic), it follows that $\langle r^2 \rangle$ is normal in $D_4$, hence $r^2$ is also in a one-point orbit (it's the only element of its order in $\langle r^2 \rangle$).
There are only two remaining elements in $\langle r \rangle$, so either they are both in the same orbit of size 2, or they are in separate one-point orbits. It now suffices to note that conjugating $r$ by $s$ gives us $r^{-1}$, so in fact $r$ and $r^{-1}$ are in the same orbit.
Summary: $\langle r \rangle$ is partitioned into three orbits by the conjugation action: namely $\{1\}, \{r^2\}$, and $\{r, r^{-1}\}$.
