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

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.