Isomorphism of group actions

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.

Solution 1:

Let $X$ and $Y$ be a left $G$-set and a left $H$-set, respectively. Then, a homomorphism of group actions is a pair $(\varphi,\sigma)$, with $\varphi:G\to H$ being a group homomorphism and $\sigma:X\to Y$ being a function compatible with the group actions in the sense that $$\sigma(g\cdot x)=\varphi(g)\cdot\sigma(x)$$ for all $g\in G$ and $x\in X$. See for example Definition 2.1 on Page 115 of Groups and Computation II by L. Finkelstein and W. M. Kantor. What you have is just an isomorphism of group actions.

If $G=H$ and $\varphi=\text{id}_G$, then $\sigma:X\to Y$ satisfying $$\sigma(g\cdot x)=g\cdot \sigma(x)$$ for all $g\in G$ and $x\in X$ is usually called a homomorphism of (left) $G$-sets. In other words, if $\sigma:X\to Y$ is a $G$-set homomorphism, then $\left(\text{id}_G,\sigma\right)$ is an example of a homomorphism of group actions from $(G,X)$ to $(G,Y)$.

Solution 2:

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.