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Consider the action of ${\rm Aut}(D_4)$ on $D_4$ by $\psi \cdot g = \psi(g)$, where $D_4$ is the dihedral group on $4$ elements (so a square). I want to find the orbits of this action.

Instead of finding the elements of ${\rm Aut}(D_4)$, I thought that maybe I could use the fact that ${\rm Aut}(D_4) \cong D_4$.

I also started out by noting that for any $\psi \in{\rm Aut}(D_4)$, $\psi(r) = r $ or $r^3$ since the order must be the same, and the remaining generator $s$ maps to $sr, sr^2, sr^3$ or , $sr^4$, but not $r^2$ because then it can't generate $D_4$. But I am stuck and would appreciate any insight.

Thanks!

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    $\begingroup$ Well, you are done ! You just have to notice that $id$, and $r^2$ are fixed by any automorphisms. You then get $4$ orbits $\{id\}, \{r^2\}, \{ r,r^3\}, \{s,sr,sr^2,sr^3\}$. $\endgroup$ – GreginGre Nov 24 '19 at 10:43
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You are practically there. You just need to compile all the information together. By the way you have defined the group action, two elements $a$ and $b$ are in the same orbit if and only if there exists an automorphism $\phi$ such that $\phi(a) = b$.

  • Note that the identity $e$ will always get sent to itself. Thus, the orbit of the identity is just $$\{e\}$$

  • Next, you have noted that $r$ can only be sent to $r$ and $r^3$ by an automorphism. Thus, the orbit of $r$ is just $$\{r,r^3\}$$ There is nothing else in this orbit, otherwise this would imply $r$ could be sent to another element besides $r$ and $r^3$.

  • Similarly, we have that the orbit of $s$ is just $$\{s, sr, sr^2, sr^3 \}$$ since you have noted that these are the only elements $s$ can be sent to.

  • Lastly, we must have that the orbit of $r^2$ is just $$\{r^2\}$$ You have already noted that $r^2$ cannot be the image of $r$ and $s$ implying it cannot be in their orbits. Also, it is not in the orbit of the identity. Thus, the only option is for $r^2$ to be in an orbit by itself. You can check that any automorphism does indeed fix $r^2$.

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