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

This fact is related to the problem of realization related to the following subgroup-defining function: Frattini subgroup
Realization problems are usually about which groups can be realized as subgroups/quotients related to a subgroup-defining function.
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This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property must also satisfy the second group property
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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

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 $Aut(G)$.