No. Take the following two actions of the Klein-4 group {e,a,b,c}:
X={a_0,a_1,b_0,b_1,c_0,c_1} where g∈{a,b,c} stabilizes g_i but swaps h_0 with h_1 for h≠g;
Y={e,a,b,c,x_0,x_1} where {e,a,b,c} is a copy of the group itself, and x_0 and x_1 are fixed by everything.
Then in each case, each nonidentity element fixes precisely 4 elements, so the characters are the same and CX≅CY.
Way I came up with this - I was trying to prove for a while that isomorphic permutation representations must come from isomorphic actions (for finite groups - not thinking about infinite groups), when I realized that this is true for a given finite group iff the characters of the transitive actions are linearly independent. But, if there are more of them than there are irreducible characters, they must be linearly dependent. I didn't see any obvious relation between the number of conjugacy classes of elements and and the number of conjugacy classes of subgroups... and it didn't look implausible that the latter could be greater sometimes (most of the time)? So, check through small groups - Klein 4 group, abelian, 4 elements, 5 subgroups. Calculate characters, find the linear dependence... and I got the example above.
Still wondering if it's true if we restrict to transitive group actions. (Still just considering finite groups, of course.)
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Date: 2007-11-27 02:18 am (UTC)X={a_0,a_1,b_0,b_1,c_0,c_1} where g∈{a,b,c} stabilizes g_i but swaps h_0 with h_1 for h≠g;
Y={e,a,b,c,x_0,x_1} where {e,a,b,c} is a copy of the group itself, and x_0 and x_1 are fixed by everything.
Then in each case, each nonidentity element fixes precisely 4 elements, so the characters are the same and CX≅CY.
Way I came up with this - I was trying to prove for a while that isomorphic permutation representations must come from isomorphic actions (for finite groups - not thinking about infinite groups), when I realized that this is true for a given finite group iff the characters of the transitive actions are linearly independent. But, if there are more of them than there are irreducible characters, they must be linearly dependent. I didn't see any obvious relation between the number of conjugacy classes of elements and and the number of conjugacy classes of subgroups... and it didn't look implausible that the latter could be greater sometimes (most of the time)? So, check through small groups - Klein 4 group, abelian, 4 elements, 5 subgroups. Calculate characters, find the linear dependence... and I got the example above.
Still wondering if it's true if we restrict to transitive group actions. (Still just considering finite groups, of course.)