Orbit of conjugation on subgroups of $D_8$

Question

Let $X$ be the set of all subgroups of $D_8$ with order $2$. For fixed $g\in D_8$, and for all $x\in X$, conjugation by $g$ is defined by
$$x\mapsto gxg^{-1}$$
What is the orbit of this group action? I have that $X=\{\{e,r^2\},\{e,s\},\{e,sr\},\{e,r^2s\},\{e,rs\}\}$

Answer 1

Orbit is that of an element of the set $X$ on which the group is acting. For example, $$\text{Orb}(\{e,r^2\})=g\{e,r^2\}g^{-1}=\{e,gr^2g^{-1}\}.$$ The second element $gr^2g^{-1}$ will run over all conjugates of $r^2$, thus you need the conjugacy class of $r^2$, which is just....... Can you take it from here?

Added explanation: Either $g=r^k$ or $g=sr^k$, with $k \in\{0,1,2,3\}$. Thus, if $g=r^k$, then the only conjugate we get is $r^kr^2r^{-k}=r^2$.

With $g=sr^k$, we get $(sr^k)r^2(sr^k)^{-1}=sr^{2}s=r^{-2}=r^2$.

So only conjugate of $r^2$ is $r^2$. This means $$\text{Orb}(\{e,r^2\})=\{\{e,r^2\}\}.$$