Prove that if A~B then Sym(A)~Sym(B).

Question

I tried to prove it with sets. Really, truly clumsy. I know |A|=|B|. Can I simply conclude that |A|!=|B|! => Sym(A)~Sym(B)?? (Sym(A) for a set A is the set of all bijections from A to A.)

Answer 1

Take a bijection $$f:A\rightarrow B.$$ Now take a permutation $\sigma \in Sym(A)$.

Show that $f\circ\sigma\circ f^{-1}$ is a permutation of $B$.

And then show that the map $$Sym(A)\rightarrow Sym(B),$$ taking $\sigma$ to $f\circ\sigma\circ f^{-1}$ is a bijection.