In particular the action of $N$ on itself must consist of $|N|$ distinct permutations. Two actions come to mind as candidates.

The regular action by left multiplication does this but it is not obvious how to extend this to an action by the larger group on $N.$ In fact this may not be possible. I considered the case that $N$ has index $2$ in the group and we set $a\cdot n=n^{-1}$ (for a chosen $a$ not in $N$) but that doesn’t quite work.

The action by conjugation does extend naturally to an action of the whole group on $N,$ but doesn’t have the desired property in all cases. It fails spectacularly for $N$ abelian. In fact if any nonidentity member of $N$ is in the center, it has the trivial action on $N.$

> If the center of $N$ consists only of the identity, then the $t \cdot n=tnt^{-1}$ has the desired property.

The thing we do not want to happen is that for some non-identity $n_1$ and $a$ we have $n_1$ and $an_1$ acting in the same way on $N$. If $n_1n_2 \neq n_2n_1$ then $a\cdot n_2 \neq (an_1) \cdot n_2.$