# The orbits with respect to two groups from the same conjugacy class are isomorphic

On the Wikipedia page for the Symbolic Method of Flajolet and Sedgewick

https://en.m.wikipedia.org/wiki/Symbolic_method_(combinatorics)

under the heading "Classes of combinatorial structures" it states: "The orbits with respect to two groups from the same conjugacy class are isomorphic."

Does anyone know of a proof of this? Perhaps something like:

Definitions/notation:

• $$G$$ is a group.

• $$X$$ is a set.

• $$\psi: G \times X \rightarrow X$$ is a $$G$$ action on $$X$$.

• $$H$$ and $$H^{\prime}$$ are conjugate subgroups of $$G$$ — that is, $$H^{\prime} = gHg^{-1}$$ for some $$g \in G$$.

• $$H$$ and $$H^{\prime}$$ act on $$X$$ by the restrictions of $$\psi$$ to $$H$$ and $$H^{\prime}$$, respectively.

• $$\text{orb}_{H}(x)$$ and $$\text{orb}_{H^{\prime}}(x)$$ denote the orbits of $$x \in X$$ with respect to $$H$$ and $$H^{\prime}$$, respectively.

• $$\phi$$ denotes the isomorphism between $$H$$ and $$H^{\prime}$$ defined by $$\phi(h) = ghg^{-1}$$.

Now, define the function $$f: \text{orb}_H(x) \rightarrow \text{orb}_{H^{\prime}}(\psi(g, x))$$ by $$\begin{equation*} f(\psi(h, x)) = \psi(\phi(h), \psi(g, x)). \end{equation*}$$ Then, $$f$$ is a well-defined bijection and \begin{align*} f(\psi(h, z)) = \psi(\phi(h), f(z)). \end{align*} Many thanks.

• What's the definition of "isomorphic orbits"? – user750041 Jun 5 '20 at 20:16
• I assume in the sense of G sets. That is, that there exists a bijection between the orbits which is compatible with the group actions. – user796754 Jun 5 '20 at 20:28
• Please do not vandalize your question, particularly after getting an answer. – Xander Henderson Jun 12 '20 at 17:45