Finite-quotient-pullbackable implies inner

Statement
Suppose $$H$$ is a finite group and $$\sigma$$ is a finite-quotient-pullbackable automorphism of $$G$$: in other words, for any surjective homomorphism $$\rho:G \to H$$, there is an automorphism $$\sigma'$$ of $$G$$ such that $$\rho \circ \sigma' = \sigma \circ \rho$$.

Then, $$\sigma$$ is an inner automorphism of $$G$$.

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 finite complete group $$G$$ such that the Fitting subgroup $$F(G)$$ is a $$p$$-group, and $$G$$ is a semidirect product of $$F(G)$$ and $$H$$. In particular, $$G/F(G) \cong H$$.

Proof
Given: A finite group $$H$$, an automorphism $$\sigma$$ of $$H$$ such that for any surjective homomorphism $$\rho:G \to H$$ there is an automorphism $$\sigma'$$ of $$G$$ such that $$\rho \circ \sigma' = \sigma \circ \rho$$.

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

Proof: Let $$p$$ be a prime not dividing the order of $$H$$. Using fact (1), there is a finite group $$G$$ such that the Fitting subgroup $$F(G)$$ is a $$p$$-group, and $$G/F(G)$$ is isomorphic to $$H$$. Let $$\rho:G \to H$$ be the quotient map.

By assumption, there exists an automorphism $$\sigma'$$ of $$G$$ such that $$\rho \circ \sigma' = \sigma \circ \rho$$. Since $$G$$ is complete, $$\sigma'$$ is inner, so there exists $$g \in G$$ such that $$\sigma'$$ is conjugation by $$g$$. It is easy to see then that $$\sigma$$ equals conjugation by the element $$\rho(g)$$, and thus, $$\sigma$$ is inner.