I saw this question in some book and I cannot manage to solve it. I wanted to try first when the order of G is finite, and just number the elements in both X and G let's say like: $$X=\left\{ x_{1},\ldots,x_{n}\right\} ,\ \ G=\left\{ g_{1},\ldots,g_{n}\right\} $$and send a pair $\left(g_{i},x_{j}\right)\mapsto x_{i}$ but it's not really a valid group action.
It feels like I am missing something simple as it just one question that leads to a bigger one later on which I did manage to solve, but somehow I am really stuck here.
Any hint is welcomed, thank you!
Be $\phi:G\to X$ one bijection between the two set, now define the action $G\times X\to X$:$$g.x=g.\phi(h):=\phi(gh). $$ Its well-defined because $h$ its uniquely determined by $x$ $(h=\phi^{-1} (x))$ and satisfy the action property: $$fg.x=fg.\phi(h)=\phi((fg)h)=\phi(f(gh))=$$ $$=f.\phi(gh)=f.g.\phi(h)=f.g.x $$ Now its easy to verify that this action its transitive and free.