This came from an old idea of mine from when I was an undergraduate for how to approach the Schreier conjecture. I (obviously) never gave it much serious thought, but wondered if much was known about it.
Let $G$ be a finite group. The group $\mathrm{Out}(G)$ has a normal subgroup $\mathrm{Out}_c(G)$ of class-preserving outer automorphisms, i.e., the group $\mathrm{Aut}_c(G)/\mathrm{Inn}(G)$, where $\mathrm{Aut}_c(G)$ is the subset of $\mathrm{Aut}(G)$ that leave invariant each conjugacy class of $G$. Note that all primes dividing $|\mathrm{Out}_c(G)|$ divide $|G|$, and if $G$ is simple then $\mathrm{Out}_c(G)=1$ (Feit-Seitz).
One approach to Schreier without CFSG is to slice $\mathrm{Out}(G)$ into two, and prove that $\mathrm{Out}_c(G)$ and $\mathrm{Out}(G)/\mathrm{Out}_c(G)$ are both soluble. Each of these looks hard, but the obvious question, even with CFSG, is:
Is $\mathrm{Out}_c(G)$ soluble for any finite group $G$?
The obvious candidates for groups with non-inner class-preserving automorphisms are $p$-groups, where this statement clearly holds. So perhaps it has a chance of being true generally. I couldn't find much progress at all on class-preserving automorphisms, so perhaps this question is too far out of reach at the moment.
See
Theorem 2.10 in this paper states the following (erratum has a correction to the original proof):
So it follows from the Schreier conjecture that $\operatorname{Out}_c(G)$ is solvable for any finite group $G$.
For infinite groups, in the paper above Sah points out the following example. Let $S$ be the symmetric group on an infinite countable set, and let $G$ be the normal subgroup formed by the permutations with finite support. In this case $\operatorname{Aut}_c(G) = S$, you can prove this the same way you prove that a transposition-preserving automorphism of the finite symmetric group is inner. Therefore $\operatorname{Out}_c(G) \cong S/G$, which is not solvable.