Finite solvable-extensible implies inner

Statement
Suppose $$H$$ is a finite solvable group and $$\sigma$$ is a finite solvable-extensible automorphism: in other words, $$\sigma$$ extends to an automorphism of $$G$$ for any finite solvable group $$G$$ containing $$H$$. Then, $$\sigma$$ is an inner automorphism of $$H$$.

Related facts

 * Finite-quotient-pullbackable implies inner
 * Hall-semidirectly extensible implies inner
 * Finite-extensible implies inner

Weaker facts

 * Finite-extensible implies class-preserving
 * Finite-extensible implies subgroup-conjugating

Facts used

 * 1) uses::Every finite group is the Fitting quotient of a p-dominated group for any prime p not dividing its order: Suppose $$H$$ is a finite group and $$p$$ is a prime not dividing the order of $$H$$. Then, there exists a p-dominated group $$G$$ with $$H$$ as Fitting quotient: in other words, there exists a finite complete group $$G$$ such that the Fitting subgroup $$F(G)$$ is a $$p$$-group, and $$H$$ is a subgroup of $$G$$ such that $$G = F(G) \rtimes H$$.
 * 2) uses::Prime power order implies nilpotent, uses::Nilpotent implies solvable
 * 3) uses::Solvability is extension-closed

Proof
Given: A finite group $$H$$, a finite solvable-extensible automorphism $$\sigma$$ of $$H$$.

To prove: $$\sigma$$ is inner.

Proof: Let $$p$$ be a prime not dividing the order of $$H$$. Consider the group $$G$$ constructed by fact (1). Note first that $$F(G)$$ is a $$p$$-group, hence by fact (2), is solvable. Thus, both $$F(G)$$ and $$G/F(G) \cong H$$ are solvable, and hence, by fact (3), $$G$$ itself is a finite solvable group.

Since $$\sigma$$ is finite solvable-extensible, $$\sigma$$ extends to an automorphism $$\sigma'$$ of $$G$$. Further, since $$G$$ is complete, there exists $$g \in G$$ such that $$\sigma'$$ is conjugation by $$g$$.

Let $$\rho:G \to H$$ be the retraction with kernel $$F(G)$$. Note that conjugation by $$g$$ preserves $$F(G)$$, hence it induces a conjugation map on $$H$$ as a quotient, namely, conjugation by the element $$\rho(g) \in H$$. However, since the restriction of $$\rho$$ to the subgroup $$H$$ is the identity map, we conclude that conjugation by $$g$$ has the same effect on $$H$$ as conjugation by $$\rho(g)$$. In particular, $$\sigma$$ equals conjugation by $$\rho(g)$$, and hence is inner.