First some motivation: most proofs that show that the group of outer automorpisms is residually finite do not actually show that the subgroup of inner automorphisms is closed in the profinite topology, they actually show something stronger: that the group of inner automorphisms is closed in the congruence topology, i.e. that a non-inner automorphism can be realised as an non-inner automorphims of some finite quotient. I would like to know whether there is a way to go back.

If $N \unlhd G$ is characteristic, the natural projection $\pi_N \colon G \to G/N$ induces a homorphism $$\tilde\pi_N \colon \mathop{Aut}(G) \to \mathop{Aut}(G/N)$$ given by $\tilde\pi_N(\phi)(gN) = \phi(g)N$ where $g \in G$ and $\phi \in \mathop{Aut}(G)$.

If $G$ is finitely generated, one can easily check that for every $K \unlhd_{f.i.} G$ there is $L \unlhd_{f.i.} G$ characteristic such that $L \leq K$, i.e. the profinite topology on $G$ is ``generated'' by characteristic subgroups of finite index.

The congruence topology on $\mathop{Aut}(G)$ is then then the topology whose base around $\mathop{id}_G$ is given by $$\mathop{Cong}(\mathop{Aut}(G)) = \{\ker{\tilde\pi_N} \mid N \unlhd_{f.i.} G \mbox{ characteristic} \}.$$

One can easily check that the congruence topology is not finer than the profinite topology on $\mathop{Aut}(G)$. In particular, if a set $\Xi \subseteq \mathop{Aut}(G)$ is closed in the congruence topology, then $\Xi$ is closed in the profinite topology.

My question is: if $G$ is a finitely generated residually finite group, does $\mathop{Inn}(G)$ being closed in the profinite topology on $\mathop{Aut}(G)$ imply $\mathop{Inn}(G)$ being closed in the congruence topology? Or equivalently, can every non-inner automorphism of $G$ be realised as an non-inner automorphism of some finite quotient of $G$ if $\mathop{Out}(G)$ is residually finite?