How do I prove that $S_A\cong S_B\implies |A|=|B|$?

Question

Let $A,B$ be infinite sets such that $S_A\cong S_B$. (Symmetric groups are group isomorphic)

How do I prove that $|A|=|B|$?

The only proof I know uses Axiom of choice. (That is, using AC to give orders to infinite family of finite sets)

Is there a way to prove this not invoking AC?