Let $G$ be an infinite group. Let $X,Y$ be two sets on which $G$ acts freely.
(An action of a group $G$ on a set $X$ is called free iff $G_x:=\{g\in G : gx=x\}$ is singleton for every $x\in X$ ).
Then how to show that there is a bijection $f: X \to Y$ such that $f(gx)=gf(x), \forall g\in G, \forall x\in X$ ?
Without additional assumptions, the claim is false: e.g. the group $\mathbb{Z}$ acts freely on both $\mathbb{Z}$ and $\mathbb{R}$ by translation, but there is no bijection between $\mathbb{Z}$ and $\mathbb{R}$ at all.
If we assume that additionally the action of $G$ is transitive, then the result is true: if $G$ acts freely and transitively on $X$ and $Y$, then there is an action-preserving bijection between $X$ and $Y$.
To prove this, fix $x\in X$ and $y\in Y$, and consider the map $$gx\mapsto gy.$$ This map:
Is well-defined, by freeness of the group action on $X$ ($g_1x=g_2x\iff g_1=g_2$).
Is defined on all of $X$, by the transitivity of the group action on $X$.
Is a surjection, by the transitivity of the group action on $Y$.
Is an injection, by the freeness of the group action on $Y$.
(A useful word to introduce at this point is "torsor".)
Alternatively, the following fact is easy to prove:
Basically, since the action is free each orbit "looks like" a copy of $G$ itself; any two orbits are therefore in action-preserving bijection, and so the only remaining possible obstacle is their number.
Note that the number of orbits can't be determined from cardinality alone: e.g. $\mathbb{Z}$ acting on $\mathbb{Z}$ by translation has one orbit, but $\mathbb{Z}$ acting on $\mathbb{Q}$ by translation has infinitely many orbits.
Note that transitivity is exactly the property of having a single orbit, so this is a generalization of the previous observation.