Equivalence of definitions of subgroup-conjugating automorphism

Subgroup-conjugating automorphism
An automorphism $$\sigma$$ of a group $$G$$ is termed a subgroup-conjugating automorphism if, for any subgroup $$H$$ of $$G$$, $$H$$ and $$\sigma(H)$$ are conjugate subgroups.

Permutation-extensible automorphism
An automorphism $$\sigma$$ of a group $$G$$ is termed a permutation-extensible automorphism if, for any injective homomorphism $$i: H \to \operatorname{Sym}(S)$$, where $$\operatorname{Sym}(S)$$ is the symmetric group on a set, there is a permutation $$\alpha$$ of $$S$$ such that $$i \circ \sigma = c_\alpha \circ i$$, where $$c_\alpha$$ denotes conjugation by $$\alpha$$.

Permutation-pushforwardable automorphism
An automorphism $$\sigma$$ of a group $$G$$ is termed a permutation-extensible automorphism if, for any homomorphism (not necessarily injective) $$\rho: H \to \operatorname{Sym}(S)$$, where $$\operatorname{Sym}(S)$$ is the symmetric group on a set, there is a permutation $$\alpha$$ of $$S$$ such that $$\rho \circ \sigma = c_\alpha \circ \rho$$, where $$c_\alpha$$ denotes conjugation by $$\alpha$$.

Applications

 * Extensible implies subgroup-conjugating

Permutation-extensible implies subgroup-conjugating
In this proof, we use the notation $$c_g$$ for conjugation by $$g$$, which is the map $$x \mapsto gxg^{-1}$$.

Given: A group $$G$$, a subgroup $$H$$, a permutation-extensible automorphism $$\sigma$$ of $$G$$.

To prove: $$\sigma(H)$$ is a conjugate subgroup to $$H$$.

Proof: The case that $$H$$ is the trivial subgroup is obvious, so we give the proof for $$H$$ nontrivial.

Let $$S = G/H$$ and $$T = G$$, with $$G$$ acting on $$S$$ and $$T$$ both by left multiplication. Let $$U$$ be the disjoint union of $$S$$ and $$T$$. $$G$$ acts faithfully on $$U$$, so we have an embedding:

$$G \to \operatorname{Sym}(U)$$.

By the condition, there exists $$\alpha \in \operatorname{Sym}(U)$$ such that $$\sigma$$ extends to conjugation by $$\alpha$$ in $$\operatorname{Sym}(U)$$. Consider the element $$H \in U$$. Clearly, the isotropy subgroup of the element $$\alpha H \in U$$ is the subgroup $$c_\alpha(H)$$ in $$G$$, which equals $$\sigma(H)$$.

Now, observe that $$\alpha H$$ cannot be in $$T$$, because then its isotropy group would be trivial, and $$\sigma(H)$$ cannot be trivial if $$H$$ is nontrivial. Thus, $$\alpha H \in S$$, so there exists $$g \in G$$ such that $$\alpha H = gH$$. Thus, the isotropy subgroup of $$\alpha(H)$$ is the conjugate subgroup $$c_g(H)$$, and thus $$\sigma(H) = c_g(H)$$.

Subgroup-conjugating implies permutation-pushforwardable
Given: A group $$G$$, an automorphism $$\sigma$$ of $$G$$ such that $$\sigma$$ sends every subgroup to a conjugate subgroup. A homomorphism $$\rho:G \to \operatorname{Sym}(S)$$ for some set $$S$$.

To prove: There exists a permutation $$\alpha$$ of $$S$$ such that $$\rho \circ \sigma = c_\alpha \circ \rho$$.

Proof: Let $$\mathcal{O}$$ be the orbit of some point in $$S$$ under the induced action of $$G$$. Let $$x \in \mathcal{O}$$, and let $$H$$ be the isotropy subgroup of $$x$$. Let $$x'$$ be any point in $$\mathcal{O}$$ whose isotropy subgroup is $$\sigma(H)$$. Such a point exists because $$H$$ and $$\sigma(H)$$ are conjugate subgroups. Now define:

$$\alpha(g \cdot x) = \sigma(g) \cdot x'$$.

This is well-defined and gives a permutation of the orbit $$\mathcal{O}$$. If we define $$\alpha$$ in this way for each orbit, we get a permutation of $$S$$ and it satisfies the condition $$\rho \circ \sigma = c_\alpha \circ \rho$$.

Puermutation-pushforwardable implies permutation-extensible
This implication is obvious.