Frattini-embedded normal-realizable implies inner-in-automorphism-Frattini

Statement
Suppose $$G$$ is a Frattini-embedded normal-realizable group: in other words, $$G$$ can be embedded in some group $$H$$ as a Frattini-embedded normal subgroup. Then $$G$$ satisfies the following condition:

The inner automorphism group of $$G$$ is a Frattini-embedded normal subgroup of the automorphism group of $$G$$.

Proof outline
Using notation as above, let $$G$$ be the group and $$H$$ be a group in which $$G$$ is embedded as a Frattini-embedded normal subgroup. Then:


 * Frattini-embedded normal is quotient-closed: Consider the map from $$H$$ to $$Aut(G)$$, sending an element of $$H$$ to its action on $$G$$ by conjugation. Let $$K$$ be the image of $$H$$. Using the fact that the property of being Frattini-embedded normal is preserved upon taking quotients, we see that $$Inn(G)$$ is a Frattini-embedded normal subgroup of $$K$$
 * Frattini-embedded normal in subgroup and normal implies Frattini-embedded normal: $$Inn(G)$$ is normal in $$Aut(G)$$, and is Frattini-embedded normal in the intermediate subgroup $$K$$, so this result tells us that $$Inn(G)$$ is Frattini-embedded normal in $$Aut(G)$$

In case $$Aut(G)$$ or $$H$$ is a finite group, we can in fact use this to conclude that $$Inn(G)$$ is contained in the Frattini subgroup of $$Aut(G)$$.