Relationship between orbits of a graph implied by different automorphisms Let $\Phi = \{\phi, \phi'\}$ be automorphisms of a finite Graph $G$. 
Suppose that $G$ has a further automorphism $\psi$. 
$\Phi.v$ is the orbit of a Node $v$ according to $\Phi$.
Under which circumstances (conditions on $\Phi$ and $\psi$) does the following hold for all nodes $v$ in the graph?
$$
\psi(\Phi.v) = \Phi.(\psi(v))
$$
If I replaced '$=$' with '$\subseteq$' or '$\supseteq$', would things change?
 A: Let $\operatorname{Aut}(G)$ be group  of all automorphisms of the graph $G$, $\overline{\Phi}\le \operatorname{Aut}(G)$ be its semigroup  generated by $\Phi$. Since the graph $G$ is finite, $\overline{\Phi}$ is a finite group. Then 
$L=\psi(\Phi.v) =\psi.\overline{\Phi}(v)$ and $R=\Phi.(\psi(v))=\overline{\Phi}.\psi(v)$. 
Thus the comparison of the sets $L$ and $R$ is equivalent to the compation of the sets $\psi^{-1}(L)=\overline{\Phi}(v)$ and $\psi^{-1}(R)=\psi^{-1}\overline{\Phi}\psi(v)$ that is of the orbits of the node $v$ by the group $\overline{\Phi}$ and its conjugated group $\psi^{-1}\overline{\Phi}\psi$. 
In particular, these orbits are equal if $\psi^{-1}\overline{\Phi}\psi=\overline{\Phi}$. The last happens exactly when both elements $\psi^{-1}\phi\psi$ and $\psi^{-1}\phi’\psi$ belong to the group $\overline{\Phi}$, in particular when $\psi$ commutes with both its generators $\phi$ and $\phi’$.
Now consider the general case. Let $S_v$ be the stablizer of the vertex $v$ that is the group of all automorphisms $\chi$ of the graph $G$ such that $\chi(v)=v$. Let $M$ be any subset of the group $\operatorname{Aut}(G)$. Then $M(v)=MS_v(v)$ and if $\xi(v)\in M(v)$ then $\xi(v)=\zeta(v)$ for some element $\zeta\in M$, so $\zeta^{-1}\xi \in S_v$ and $\xi\in MS_v$. Thus 
$$MS_v=\{\xi\in \operatorname{Aut}(G):\xi(v)\in M\}.$$ 
Hence $\overline{\Phi}(v)$ $=$ (resp., $\subseteq$, $\supseteq$) $\psi^{-1}\overline{\Phi}\psi(v)$ iff 
$\overline{\Phi}S_v$ $=$ (resp., $\subseteq$, $\supseteq$) $\psi^{-1}\overline{\Phi}\psi S_v$.
