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.)

According to the Schreier–Ulam–Baer theorem, the nontrivial normal subgroups of $S(X)$ are (i) the subgroup $S_\mathrm{fin}(X)$ of permutations of $X$ of finite support, (ii) the subgroup $A_\mathrm{fin}(X)$ of $S_\mathrm{fin}(X)$ of even permutations, and (iii) for each cardinal $\kappa$ the subgroups $S_{<\beta}(X)$ and $S_{\leq\beta}(X)$ of permutations which move strictly less than $\beta$ points and at most $\beta$ points, respectively.

Since, as you said, a cardinal is determined by the order type of the set of cardinals below it, looking at the lattice of normal subgroups of $S(X)$, then, lets you guess the cardinal of $X$.
