IA equals inner in extraspecial

Statement
In an fact about::extraspecial group (a group of prime power order whose commutator subgroup, center and Frattini subgroup are all equal and cyclic) any fact about::IA-automorphism is inner.

Extraspecial group
A group of prime power order is termed an extraspecial group if its commutator subgroup, center, and Frattini subgroup are all equal, and this subgroup is a cyclic group of prime order.

IA-automorphism
An IA-automorphism of a group is an automorphism that induces the identity automorphism on the Abelianization of the group (the quotient by its commutator subgroup).

An equivalent fact

 * Extraspecial implies inner automorphism group is self-centralizing in automorphism group

Proof
Given: An extraspecial $$p$$-group $$G$$. Let $$A$$ denote the group of IA-automorphisms of $$G$$, and $$I$$ denote the group of inner automorphisms of $$G$$

To prove: $$I = A$$

Proof: Note that $$I \le A$$ (i.e., any inner automorphisms is IA). Also, $$I \cong G/Z(G)$$ (again, a standard fact). Since both groups are finite, it suffices to show that $$|A| \le |I|$$.

The first step in this is to show that elements of $$A$$ act as the identity. not only on the quotient $$G/G'$$, but also on $$G' = Z(G)$$. Thus, $$A$$ can be viewed as the stability group of the series $$1 \le Z(G) \le G$$. So, what we need to prove is that the cardinality of this stability group is at most $$|I| = |G/Z(G)|$$.