The condition that $N$ is normal isn’t important, see the end. I start with an analysis that uses the structure of $N$:

 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 has this property, but it is not obvious how to extend this to an action by $G$ on $N.$ In fact we have seen that this may not be possible. It is for the cyclic subgroup of order $3$ in $S_3.$ 

The action by conjugation does extend naturally to an action of $G$ on $N,$ (as $N$ is normal in $G$) 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 $a$ and $an_1$ acting in the same way on $N$. If $n_1n_2 \neq n_2n_1$ then, for this action, $a\cdot n_2 \neq (an_1) \cdot n_2.$

> Suppose merely that $G$ has a representation acting faithfully on a set of size $k$ and a subgroup (maybe not normal) of order $k.$ Arbitrarily identify the members of the set with the elements of $N.$ Then your desired condition holds for all of $G$, not merely the cosets of $N.$