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.
After defining conjugacy as a group action, it follows from the definition that the orbits are now the conjugacy classes.