$G$-set isomorphisms for free action on some product Let $G$ be a finite group, $X$ a finite set on which $G$ acts freely. Consider the set $G\times X$ with diagonal action of $G$, i.e.
$ h.(g,x) = (hg, h.x).$
This is also free. For the orbits, we have $$G(g,x)=G(e,g^{-1}.x)\cong G.$$ From Burnside's Lemma and freeness we get the number of orbits:
$$ |(G\times X)/G| = |G\times X|/|G| = |X|.$$
On the other hand, let ${}_\epsilon X$ be the same set but with trivial $G$-action. The orbits in $G \times{}_\epsilon X$ are again isomorphic to $G$, and indexed by $x\in X$ (of the form $G(e,x)$), hence
$$ |G\times{}_\epsilon X|/|G| = |X|$$
as well.
So both $G\times X$ and $G\times{}_\epsilon X$ are isomorphic to $$\bigsqcup_{x\in X}G.$$
I have been told that these are even $G$-set isomorphisms, but I can't quite see it. My questions are now:


*

*What is the action on the coproduct?


My initial guess was $h.g_x = (hg)_x$, but this doesn't work.



*If 1. doesn't help me: What are the isomorphisms?


*Lastly, can we say something similar if the action of $G$ on $X$ isn't free?

Thanks for any input.

Further thoughts: I guess the disjoint union may have different actions, depending  on which of the products induces it. But then we have two set-theoretically isomorphic $G$-sets which are not $G$-isomorphic, which contradicts the known true statement I want to show:
$$ G\times X \cong_{G\mathrm{-set}}G\times{}_\epsilon X $$
 A: Let us clear this up. We can write
$$ G\times X = \bigsqcup_{x\in X} G(e,x)\cong \bigsqcup_{x\in X} G_{|x}, $$
where on the RHS the $|x$ is meant as an index. Why can we do that? Because if $x\neq y$, then the orbits $G(e,x)$ and $G(e,y)$ are disjoint. The isomorphism, let's call it $\phi$, is then given [on the "orbit generators"] by $(e,x)\mapsto e_{|x}$, where $G$ acts on the RHS via $g.{e_{|x}} = g_{|x}$. We want this action because the orbits of a free action can be seen as copies of the regular action. (Of course, $g.h_{|x} \equiv (hg^{-1})_{|x}$ would also have been possible, but why make life more complicated than it already is?)
$\phi$ is extended as a $G$-set morphism, i.e. 
$$\phi(g,g.x)=\phi(g.(e,x))\equiv g.(\phi(e,x)) = g_{|x}.$$
For a general element $(g,x)$ we thus have
$$\phi(g, x) = \phi(g.(e,g^{-1}.x))= g_{|g^{-1}.x}. $$
It is important to stress here that $g^{-1}.x$ is a point coming from the fixed free action of $G$ on $X$. The inverse works much the same way, $\phi^{-1}(g_{|g^{-1}.x}) = \phi^{-1}(g.e_{|g^{-1}.x}) = g.\phi^{-1}(e_{|g^{-1}.x}) = g.(e,g^{-1}.x)=  (g,x)$, as expected.
For the part with the trivial action, similar arguments are easily made, but the isomorphism, which we will call $\psi: G\times\,_\epsilon X \to \bigsqcup G$ is given simple by $\psi(g, x) = g_{|x}$. 
Okay, now since $G\times X$ and $G\times\,_\epsilon X$ are $G$-set isomorphic to the same set, they are themselves isomorphic as $G$-sets, via $\phi^{-1}\circ \psi$ and $\psi^{-1}\circ \phi$. Let us give the isomorphisms explicitly:
\begin{align*}
\phi^{-1}\circ \psi:G\times\,_\epsilon X &\to G\times X\\
(g,x) &\mapsto \phi^{-1} (g_{|x})= (g,g.x) \\[2em]
\psi^{-1}\circ \phi:G\times X &\to G\times\,_\epsilon X\\
(g,x) &\mapsto \phi^{-1} (g_{|g^{-1}.x})= (g,g^{-1}.x) \\[2em]
\end{align*} 

A final note: This is the motivation for the following. Given a Hopf algebra $H$ over $k$, a $H$-module $V$, and the same module equipped with the trivial action, $\,_\epsilon V$, we have the $H$-module isomorphism 
\begin{align*}
H\otimes_k\,_\epsilon V &\xrightarrow{\sim} H\otimes_k V\\
h\otimes v&\mapsto h_1\otimes h_2.v\ ,
\end{align*} 
with inverse $h\otimes v \mapsto h_1\otimes S(h_2).v$. Indeed,
\begin{align*}
h\otimes v \mapsto h_1\otimes h_2.v&\mapsto h_{1_1}\otimes S(h_{1_2})h_2.v \\&=h_{1}\otimes S(h_{2_1})h_{2_2}.v \\
&=h_{1}\otimes \epsilon(h_2)v \\
&= h\otimes v
\end{align*} 
Note that in the case of a group Hopf algebra $kG$, these maps are pretty much the same as above for the $G$-set case: $S(g) = g^{-1}$ for all $g\in G$.
