Fixed-point subgroup of a subgroup of the automorphism group implies local powering-invariant

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., fixed-point subgroup of a subgroup of the automorphism group) must also satisfy the second subgroup property (i.e., local powering-invariant subgroup)
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Statement

Suppose $G$ is a group. Suppose $B$ is a subgroup of the automorphism group of $G$. Suppose $H$ is the subgroup of $G$ comprising precisely those elements that are fixed by every element of $B$. In other words, $H$ is a fixed-point subgroup of a subgroup of the automorphism group in $G$.

Then, $H$ is a local powering-invariant subgroup of $G$: if $h \in H$ and $n \in \mathbb{N}$ are such that there is a unique $x \in G$ satisfying $x^n = h$, then $x \in H$.

Proof

Given: Group $G$, subgroup $B$ of $\operatorname{Aut}(G)$. Subgroup $H$ of $G$ defined as the set of fixed points under $B$ of $G$. $h \in H$ and $n \in \mathbb{N}$ are such that there is a unique $x \in G$ satisfying $x^n = h$.

To prove: $x \in H$. In other words, $\sigma(x) = x$ for all $\sigma \in B$.

Proof: We do the proof for fixed but arbitrary $\sigma \in B$. We have that, since $\sigma$ is an automorphism: $(\sigma(x))^n = \sigma(x^n)$

Simplifying further: $(\sigma(x))^n = \sigma(x^n) = \sigma(h) = h$

where the last step follows from the fact that $h \in H$ and every element of $H$ is fixed by every automorphism in $B$.

We thus obtain that $(\sigma(x))^n = h$. Since $x$ is the unique element whose $n^{th}$ power is $h$, this forces $\sigma(x) = x$, completing the proof.