Finding orbit in a group action of automorphism group of dihedral group on the dihedral group.
Let $$D_{2n}=\langle a,b:a^n=b^2=1,bab^{-1}=a^{-1}\rangle $$ be the dihedral group of order $2n$ and $Aut(D_{2n})=\{\beta_{k,j}:1\leq k,j\leq n-1, (k,n)=1\}$ be its automorphism group such that $\beta_{k,j}$ is the unique automorphism on $D_{2n}$ induced by the map $a\to a^k$ and $b\to a^jb$. Consider the group action $\phi:Aut(D_{2n})\times D_{2n}\to D_{2n}$ such that $\phi(\alpha,g)=\alpha(g)$ for all $g\in D_{2n}$ and $\alpha\in Aut(D_{2n})$.
If $n$ is odd then $$ orb(a)=\{\beta_{k,j}(a):\beta_{k,j}\in Aut(G) ~and~~ 1\leq k,~j\leq n-1,~~ (k,n)=1\}=\{a^k:1\leq k\leq (n-1),~(n,k)=1\}. $$ Similarly, $$orb(a^2)=orb(a^3)=\dots = orb(a^{n-1})=\{a^k:1\leq k\leq (n-1),~(n,k)=1\}.$$
In general, can we find $orb(a),~orb(a^2),~orb(a^3),~\dots,~orb(a^{n-1})$ when $n$ is even? In particular, we have obtained the following orbits.
If $n=4$ then $orb(a)=\{a,~a^3\}=orb(a^3)$, $orb(a^2)=\{a^2\}$.
If $n=6$ then $orb(a)=\{a,~a^5\}=orb(a^5)$, $orb(a^2)=\{a^2,~a^4\}=orb(a^4)$ and $orb(a^3)=\{a^3\}$.
If $n=8$ then $orb(a)=\{a,~a^3,~a^5,~a^7\}=orb(a^3)=orb(a^5)=orb(a^7)$, $orb(a^2)=\{a^2,~a^6\}=orb(a^6)$ and $orb(a^4)=\{a^4\}$.
If $n=10$ then $orb(a)=\{a,~a^3,~a^7,~a^9\}=orb(a^3)=orb(a^7)=orb(a^9)$, $orb(a^2)=\{a^2,~a^4,~a^6,~a^8\}=orb(a^4)=orb(a^6)=orb(a^8)$ and $orb(a^5)=\{a^5\}$.