Can the symmetric groups on sets of different cardinalities be isomorphic?

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 $X$.

The following seems simpler than the answers given earlier. I apologize if this answer (or a simpler one) is already in one of the links and I overlooked it.

I claim that, for any infinite set $X$, the cardinality $|X|$ can be obtained from the symmetric group Sym$(X)$ as the smallest cardinality of any conjugacy class other than the trivial class $\{1\}$. First, there is a conjugacy class of size $|X|$, for example, the class of those permutations that just interchange two elements of $X$ while fixing everything else.

Now the main point: Suppose $C$ is a non-trivial conjugacy class. Consider any element $\pi$ of $C$ and any element $x\in X$ moved by $\pi$. (Such an $x$ exists as $C$ isn't the trivial class.) For any element $y\in X-\{x\}$, consider the permutation that sends $\pi(x)$ and $y$ to each other and fixes everything else (including, in particular, $x$). Then $\sigma\pi\sigma^{-1}$ (also known as $\sigma\pi\sigma$) is a conjugate of $\pi$ that sends $x$ to $y$. So the $|X|$ different possible $y$'s give us $|X|$ different conjugates of $\pi$. Therefore, $|C|\geq|X|$, as claimed.

Remark: I used the Axiom of Choice twice here, first to say that the number of pairs from $X$ is $|X|$, and second to say that the number of possible $y$'s is also $|X|$. (The second fact follows easily from the first.) I don't know whether the result holds in the absence of the Axiom of Choice.

In an ancient paper with Saharon Shelah, I proved that if κ < λ, then Sym(λ) does not embed into Sym(κ). The proof is based on results in an even more ancient paper with John Dixon and Peter Neumann. The relevant papers are:

Saharon Shelah and Simon Thomas, Implausible subgroups of infinite symmetric groups. Bull. London Math. Soc. 20 (1988), no. 4, 313--318.

John D. Dixon, Peter M. Neumann and Simon Thomas, Subgroups of small index in infinite symmetric groups. Bull. London Math. Soc. 18 (1986), no. 6, 580--586.