Conjugation actions between subgroups that normalize each other are compatible

Statement
Suppose $$G$$ and $$H$$ are subgroups of a group $$Q$$ such that each subgroup normalizes the other. More explicitly, $$G$$ is contained in the normalizer $$N_Q(H)$$ and $$H$$ is contained in the normalizer $$N_Q(G)$$. Then, define the following actions of the groups on each other:


 * $$\alpha:G \to \operatorname{Aut}(H)$$ is defined as follows: for $$g \in G$$, $$\alpha(g)$$ is the automorphism $$h \mapsto ghg^{-1}$$.
 * $$\beta:H \to \operatorname{Aut}(G)$$ is defined as follows: for $$h \in H$$, $$\beta(h)$$ is the automorphism $$g \mapsto hgh^{-1}$$.

Then, $$\alpha$$ and $$\beta$$ form a compatible pair of actions between $$G$$ and $$H$$.

Proof
Denote by $$\cdot$$ each of the actions of the groups on each other (via $$\alpha$$ or $$\beta$$) and each action of a group on itself (via conjugation). Note that the notation is unambiguous, because all the actions arise as restrictions of conjugation in $$Q$$, so that where elements live in both $$G$$ and $$H$$, the actions all coincide.

To prove: $$g_1 \cdot (h \cdot g_2) = (g_1 \cdot h) \cdot (g_1 \cdot g_2) \ \forall \ g_1,g_2 \in G, h \in H$$

$$h_1 \cdot (g \cdot h_2) = (h_1 \cdot g) \cdot (h_1 \cdot h_2) \ \forall g \in G, h_1,h_2 \in H$$

Proof: The crude proof is below, but a more sophisticated derivation follows from noting that these follow from the concept of conjugation rack of a group.

First equality, left side: $$g_1 \cdot (h \cdot g_2) = g_1(hg_2h^{-1}g_1^{-1} = g_1hg_2h^{-1}g_1^{-1}$$

First equality, right side: $$(g_1 \cdot h) \cdot (g_1 \cdot g_2) = (g_1hg_1^{-1}) \cdot (g_1g_2g_1^{-1}) = (g_1hg_1^{-1})(g_1g_2g_1^{-1})(g_1hg_1^{-1})^{-1} = g_1hg_2g_1^{-1}(g_1h^{-1}g_1^{-1}) = g_1hg_2h^{-1}g_1^{-1}$$

Second equality, left side: $$h_1 \cdot (g \cdot h_2) = h_1(gh_2g^{-1})h_1^{-1} = h_1gh_2g^{-1}h_1^{-1}$$

Second equality, right side: $$(h_1 \cdot g) \cdot (h_1 \cdot h_2) = (h_1gh_1^{-1}) \cdot (h_1h_2h_1^{-1}) = (h_1gh_1^{-1})(h_1h_2h_1^{-1})(h_1gh_1^{-1})^{-1} = h_1gh_1h_1^{-1}h_1g^{-1}h_1^{-1} = h_1gh_2g^{-1}h_1^{-1}$$