For any set X, let SX be the symmetric group on X, the group of permutations of X.
My question is: Can there be two nonempty sets X and Y with different cardinalities, but for which SX is isomorphic to SY?
Certainly there are no finite examples, since the symmetric group on n elements has n! many elements, so the finite symmetric groups are distinguished by their size.
But one cannot make such an easy argument in the infinite case, since the size of SX is 2|X|, and the exponential function in cardinal arithmetic is not necessarily one-to-one.
Nevertheless, in some set-theoretic contexts, we can still make the easy argument. For example, if the Generalized Continuum Hypothesis holds, then the answer to the question is No, for the same reason as in the finite case, since the infinite symmetric groups will be characterized by their size. More generally, if κ < λ implies 2κ < 2λ for all cardinals, (in another words, if the exponential function is one-to-one, a weakening of the GCH), then again Sκ is not isomorphic to Sλ since they have different cardinalities. Thus, a negative answer to the question is consistent with ZFC.
But it is known to be consistent with ZFC that 2κ = 2λ for some cardinals κ < λ. In this case, we will have two different cardinals κ < λ, whose corresponding symmetric groups Sκ and Sλ nevertheless have the same cardinality. But can we still distinguish these groups as groups in some other (presumably more group-theoretic) manner?
The smallest instance of this phenomenon occurs under Martin's Axiom plus ¬CH, which implies 2ω = 2ω1. But also, if one just forces ¬CH by adding Cohen reals over a model of GCH, then again 2ω = 2ω1.
(I am primarily interested in what happens with AC. But if there is a curious or weird counterexample involving ¬AC, that could also be interesting.)