No: For any infinite set $X$, the cardinality of $S(X)$ is the same as that of its power set $\mathcal P(X)$ and therefore (by Cantor's theorem) different from the cardinality of $X$ itself.

We need to know that there is a bijection $f: X\times X\to X$. That this exists for every infinite $X$ is a well-known consequence (and, indeed, equivalent) of the axiom of choice.

Now in one direction, every permutation $\sigma\in S(X)$ can be encoded as a subset of $X$, namely $\{ f(x,\sigma(x)) \mid x\in X \}$, so there are at least as many subsets as there are permutations.

In the other direction: Choose two different $x,y\in X$. Then we can encode any subset $A\subseteq X$ as a permutation of $X$, namely the one that swaps $f(x,a)$ and $f(y,a)$ for all $a\in A$, and leaves all other elements of $X$ unchanged. This shows there are at least as many permutations as there are subsets.

The Cantor-Bernstein theorem now tells us that $S(X)$ and $\mathcal P(X)$ are equinumerous.