Let $X$ and $Y$ be two sets s.t. $|X|=|Y|.$ Show that the groups $\operatorname{Sym}(X)$ and $\operatorname{Sym}(Y)$ of all permutations of $X$ and $Y$, respectively, are isomorphic.
My attempt: Since $|X|=|Y|$ we can assume there exists a bijection $\phi:X\rightarrow{Y}$. I remember successfully trying this question before with success by defining $\psi:\operatorname{Sym}(X)\rightarrow{\operatorname{Sym}(Y)},$ defined by $\psi(\sigma):=\phi{}\circ\sigma\circ\phi^{-1}$, but can't quite remember its use.
Note: I am trying to do this with basic assumed knowledge on abstract algebra. Also I am aware there are two other questions that have answers with very similar questions, but I would like a more simplistic approach.
The map $\psi$ is a map from $\operatorname{Sym}(X)$ into $\operatorname{Sym}(Y)$. It is easy to check that it is a group homomorphism. Besides, if you define $\eta\colon\operatorname{Sym}(Y)\longrightarrow\operatorname{Sym}(X)$ by $\eta(\sigma)=\phi^{-1}\circ\sigma\circ\phi$, then $\eta$ is clarly the inverse of $\psi$, besides being a group homomorphism too. Therefore, $\psi$ is actually a group isomorphism.