TITLE: Isomorphism of group actions
QUESTION [6 upvotes]: Let $G$ be a group acting on a set $X$, and let $H$ be a group acting on a set $Y$. I'm wondering if there is standard terminology or notation to express the following condition: there exist an isomorphism $\varphi:G\rightarrow H$ and a bijection $\sigma:X\rightarrow Y$ such that $$\sigma(gx)=\varphi(g)\sigma(x)$$
for all $g\in G$ and all $x\in X$. Basically, I want to say that not only are $G$ and $H$ isomorphic as abstract groups, but that they act in the same way on the corresponding sets.

REPLY [5 votes]: An action of $G$ on $X$ is given by a group homomorphism $G \to Sym(X)$. 
Therefore, the action of $G$ on $X$ is isomorphic to the action of $H$ on $Y$ exactly when the diagram below commutes:
$$
\begin{array}{ccc}
G & \to & Sym(X) \\
\downarrow & & \downarrow \\
H & \to & Sym(Y)
\end{array}$$
The horizontal arrows are induced by the actions, the vertical arrow on the left is $\varphi$, and the vertical arrow on the right is induced by $\sigma$. 
The diagram can be used to define morphisms of group actions, not just isomorphisms.