KGS math club/solution 11 15

D239500800 (06/04/12 17:21): One may take N=6 and G = H = S_6. S_6 can act transitively on 6 points the usual way, or by acting by conjugation on the 6 transitive subgroups of S_6 which are isomorphic to S_5. One way to obtain a transitive S_5 on 6 points is to consider the group of permutations of the vertices of a Rubik's Cube generated by rotating two adjacent faces.

Another solution: S_4 acting on 6 items. If the 6 are the vertices of an icosahedron, half the permutations are odd. If the 6 are the pairs of antipodal edges of an icosahedron, all the permutations are even. So the two group actions cannot be isomorphic.