Injectivity of the functor from group actions to permutation representations

Question

Let $G$ be a finite group acting on a finite set $X$. We can take the free $\mathbb{C}$-vector space generated by $X$ which we can denote $V_X$. We get a representation of $G$ on $V_X$ coming from the action on $X$. This defines a functor from $G$-sets to $G$-representations. Does this functor fail to be injective? Namely, can we write down an example of $G$-sets $X$ and $Y$ that are not isomorphic but whose associated representations are isomorphic?

Answer 1

Transitive $G$-sets up to isomorphism (i.e. $G$-equivariant bijection) are in one to one correspondence with subgroups $H \subset G$ up to conjugacy. The correspondence just sends a transitive $G$-set to the stabilizer of a point, which only depends on the choice of point up to conjugacy.

On the other hand, up to isomorphism there are the same number of irreducible complex representations as conjugacy classes in $G$.

So in particular this map can't possibly be injective provided there are more conjugacy classes of subgroups than of elements, which seems to be the norm provided the group is large enough and nonabelian. For example $S_4$ has $4$ conjugacy classes of elements, but at least $5$ conjugacy classes of subgroups: The four cyclic ones generated by a single element, plus alternating group $A_4$. (as pointed out in the comments, in fact there are 11 subgroups up to conjugacy)