Let $N$ be a normal subgroup of $G$. $G$ acts from right on the set $$R_m=\left\{\rho: N \longrightarrow G L_m(F)\;|\;\quad \rho \text{ is a group homo.}\right\}$$ as follows: for $g \in G,(\rho \cdot g)(n):=\rho\left(g n g^{-1}\right)$ for every $n \in N$.
Question: This action, also defines an action of $G / N$ on $R_m$.
Attempt:
Now I want to use universal property of quotient groups and I want to use the following version of an $G$ action on $R_m$. $$a:G\to Aut(R_m)$$
and
$$ \require{AMScd} \begin{CD} G @>\pi_N>> G/N\\ @. {_{\rlap{\ a}}\style{display: inline-block; transform: rotate(31deg)}{{\xrightarrow[\rule{4em}{0em}]{}}}} @VV\exists ! \bar aV\\ @. Aut(R_m) \end{CD}$$
To be able to say existence of such $\bar a$ group homomorphism, I need to show $ker(a)\supset N$
So the group homomorphism $a$ is actually defined as:
$$a(g):R_m\to R_m\\ \rho\mapsto \rho\cdot g$$
So since the identity element of $Aut(R_m)$ is identity. I need to show for given $n_0\in N$ the map $a(n_0)=Id_{R_m}$
So equivalently need to show for every $n\in N$: $$a(n_0)(\rho)(n)=(\rho\cdot n_0)(n)=\rho(n_0nn_0^{-1})\\=\rho(n)$$
But I cannot recall if $N\subset G$ is a normal, these conjugates commute? Do we have $n,n_0\in N$ , $n_0nn_0^{-1}=n$?