Tensor product of groups is commutative up to natural isomorphism

Statement
Suppose $$G$$ and $$H$$ are (not necessarily abelian) groups with a compatible pair of actions $$\alpha:G \to \operatorname{Aut}(H)$$ and $$\beta:H \to \operatorname{Aut}(G)$$. We can define the tensor product of groups $$G \otimes H$$ and also the tensor product of groups $$H \otimes G$$. The claim is that these tensor products are isomorphic groups with a natural isomorphism $$G \otimes H \to H \otimes G$$ given as follows on a generating set:

$$g \otimes h \mapsto (h \otimes g)^{-1}$$

Note that this natural isomorphism is self-inverse, i.e., the natural isomorphism from $$G \otimes H$$ to $$H \otimes G$$ is the inverse map to the similarly defined natural isomorphism from $$H \otimes G$$ to $$G \otimes H$$.

Related facts

 * Tensor product of Lie rings is commutative up to natural isomorphism

Proof idea
Consider the two axioms that define the tensor product:


 * $$(g_1g_2) \otimes h = ((g_1 \cdot g_2) \otimes (g_1 \cdot h))(g_1 \otimes h)$$
 * $$g \otimes (h_1h_2) = (g \otimes h_1)((h_1 \cdot g) \otimes (h_1 \cdot h_2))$$

Note that apart from interchanging the roles of $$G$$ and $$H$$, the other key difference between the two axioms is that the order of multiplication on the respective right sides differs. Since the inverse map is involutive, this explains why $$g \otimes h \mapsto (h \otimes g)^{-1}$$ works.