There are many group actions on sets which are linearly equivalent but not equivalent as actions. In fact, every group other than the cyclic group has one. This follows from some easy linear algebra:
- the number of irreducible reps over $\mathbb{Q}$ is the number of conjugacy classes of cyclic subgroups of $G$, (EDIT:there are at most this many since any two elements which generate conjugate cyclic groups have the same character in a rational representation; on the other hand, the characters of the inductions of the trivial from any set of cyclic groups, no two of which are conjugate, are linearly independent, so there are at least this many) and
- the number of non-isomorphic transitive G-sets is the number of conjugacy classes of subgroups.
Thus, there must be an integer valued linear combination of transitive actions which has trivial character. Moving all the actions with negative coefficients to the other side of the equality, we get two different actions with the same character, and thus isomorphic representations.
I actually wrote a paper about this a few years back, which I think is a reasonable starting place for the subject, which actually has quite a long history, and a reasonably extensive literature.
