Left-transitively permutable implies characteristic

Statement
Suppose $$H$$ is a subgroup of a group $$K$$ such that whenever $$K$$ is a fact about::permutable subgroup of a group $$G$$, $$H$$ is also a permutable subgroup of $$G$$. Then, $$H$$ is a characteristic subgroup of $$K$$.

Facts used

 * 1) uses::Every group is normal fully normalized in its holomorph

Proof
Given: A subgroup $$H$$ of a group $$K$$ such that whenever $$K$$ is a permutable subgroup of a group $$G$$, then $$H$$ is permutable in $$G$$.

To prove: $$H$$ is characteristic in $$K$$: for any automorphism $$\sigma$$ of $$K$$, and any $$g \in H$$, $$\sigma(g) \in H$$.

Proof: Let $$G$$ be the holomorph of $$K$$; in other words, we have:

$$G = K \rtimes \operatorname{Aut}(K)$$.

Let $$\varphi:G \to \operatorname{Aut}(K)$$ be the natural retraction with kernel $$K$$.

Now, consider $$\sigma$$ as an element of $$G$$ (via its membership in $$\operatorname{Aut}(K)$$, and $$g$$ also as an element of $$G$$. Let $$A$$ be the cyclic subgroup of $$G$$ generated by $$\sigma$$. Since by assumption $$H$$ is permutable in $$G$$, we have:

$$AH = HA$$

In particular, consider the product $$\sigma g$$. This is in $$AH$$, hence it is also in $$HA$$, so there exists $$h \in H, \alpha \in A$$ such that $$\sigma g = h \alpha$$. Applying $$\varphi$$ to both sides, we get:

$$\varphi(\sigma g) = \varphi(h \alpha)$$

This yields:

$$\sigma = \alpha$$.

In particular, we obtain that:

$$\sigma g = h \sigma$$.

Thus, we obtain that:

$$\sigma g \sigma^{-1} = h \in H$$

Since the action of $$\operatorname{Aut}(K)$$ is by conjugation, this yields:

$$\sigma(g) = \sigma g \sigma^{-1} \in H$$.