P-group of automorphisms of p-group is contained in stability group of some normal series

Statement
Suppose $$G$$ is a finite $$p$$-group, (i.e., a group of prime power order), and $$H \le \operatorname{Aut}(G)$$ is also a $$p$$-group. Then, we can find a normal series of $$G$$:

$$\{ e \} = N_0 \le N_1 \le N_2 \le \dots \le N_r = G$$

where each $$N_i$$ is normal in $$G$$, and where the induced action of $$H$$ on each subquotient $$N_i/N_{i-1}$$ is trivial. In other words, $$H$$ is contained in the stability group of this normal series.

Particular cases
A particular case of interest is where $$H = \operatorname{Inn}(G)$$. In this case, we can take the upper central series of $$G$$ (the procedure outlined below does yield the upper central series). We can also take the lower central series of $$G$$, or in fact any central series of $$G$$.

Facts used

 * 1) Nilpotent implies center is normality-large

Proof
Given: A $$p$$-group $$G$$, a $$p$$-group $$H \le \operatorname{Aut}(G)$$

To prove: There exists a normal series for $$G$$ such that $$H$$ is in the stability group of the normal series.

Proof: We prove the statement by induction on the order of $$G$$.

Consider the semidirect product $$K = G \rtimes H$$. This is a $$p$$-group, and $$G$$ is a normal subgroup. By fact (1), $$G$$ intersects the center of $$K$$ nontrivially. Let $$N$$ be this intersection. Then, $$N$$ is normal in $$K$$ (because it is the intersection of two normal subgroups) and hence also in $$G$$.

Now, consider the induced action of $$H$$ on $$G_1 = G/N$$. This induced action gives a homomorphism $$H \to \operatorname{Aut}(G/N)$$. Let $$H_1$$ be the image of $$H$$. Then, we have a strictly smaller $$p$$-group $$G_1$$ and a $$p$$-group $$H_1 \le \operatorname{Aut}(G_1)$$. By induction, there exists a normal series for $$G_1$$ such that $$H_1$$ acts as stability automorphisms for it. Taking the inverse image of this series under the quotient map $$G \to G/N$$ yields a normal series of $$G$$ such that $$H$$ acts as stability automorphisms of it.