Question about isomorphism between the subgroups $N$ and $a^{-1}Na$

Question

In showing isomorphism between the subgroups $N$ and $a^{-1}Na$ of a group $G$, one usually define a function $f:N\rightarrow a^{-1}Na$ by $f(n)=a^{-1}na$ for all $a\in G$ and some $n\in N$. This means that $f(N)=a^{-1}Na$ since there is a bijection between $N$ and $a^{-1}Na$, but does this imply $N=a^{-1}Na$ which would mean that $N$ is a normal subgroup of $G$?

Thank you in advance.

Answer 1

Having a bijection means that the subgroups are isomorphic, but they don't have to have the same elements. Let $G$ be a group of all permutations on 3 elements 1,2,3, $N$ be a subgroup of permutations which keep element 1 in place, and $a$ be some permutation which moves 2 to 1. In this situation $a^{-1} N a$ is a subgroup of all permutations which keep 2 in place, which is different from $N$, which keeps 1 in place.

Answer 2

No.

Consider $N=\{e,(12)\}\le S_3$ and $a=(13)$. Then $N\neq a^{-1}Na$.