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Remark. One thing I still find weird about this is that it almost works to give a homomorphism $S_n \to S_3$ as well, by taking $\operatorname{Inj}(3,n)$ instead of $\operatorname{Inj}(2,n)$. It somehow only fails because $S_3$ is not abelian.

Remark. One thing I still find weird about this is that it almost works to give a homomorphism $S_n \to S_3$ as well, by taking $\operatorname{Inj}(3,n)$ instead of $\operatorname{Inj}(2,n)$. It somehow only fails because $S_3$ is not abelian. Replacing $S_3$ by its subgroup $C_3$ does give a homomorphism $S_n \to C_3$, which presumably is trivial. But somehow for $C_2 = S_2$ the map is nontrivial!

In the case of $S_n \times S_2$ $\style{display: inline-block; transform: rotate(90deg)}{\circlearrowright}$ $\!\operatorname{Inj}(2,n)$, this gives a homomorphism $S_n \to S_2$. Given a section $s \colon {T \choose 2} \to T^2 \setminus \Delta_T$, write $s_1, s_2 \colon {T \choose 2} \to T$ for the projections $\pi_i \circ s$. Then the crossed homomorphism $\phi \colon S_n \to \{\pm 1\}^Y$ can be identified with $$\phi(\sigma)(S) = \frac{x_{\sigma(s_1(S))}-x_{\sigma(s_2(S))}}{x_{s_1(\sigma(S))}-x_{s_2(\sigma(S))}}.$$ Taking the product over all $S \in {T \choose 2}$ recovers the formula in Speyer's answer when $s$ is given by $S \mapsto (\min(S),\max(S))$.

In the case of $S_n \times S_2$ $\style{display: inline-block; transform: rotate(90deg)}{\circlearrowright}$ $\!\operatorname{Inj}(2,n)$, this gives a homomorphism $S_n \to S_2$. Given a section $s \colon {T \choose 2} \to T^2 \setminus \Delta_T$, write $s_1, s_2 \colon {T \choose 2} \to T$ for the projections $\pi_i \circ s$. Then the crossed homomorphism $\phi \colon S_n \to \{\pm 1\}^{T \choose 2}$ can be identified with $$\phi(\sigma)(S) = \frac{x_{\sigma(s_1(S))}-x_{\sigma(s_2(S))}}{x_{s_1(\sigma(S))}-x_{s_2(\sigma(S))}}.$$ Taking the product over all $S \in {T \choose 2}$ recovers the formula in Speyer's answer when $s$ is given by $S \mapsto (\min(S),\max(S))$.

Remark. One thing I still find weird about this is that it almost works to give a homomorphism $S_n \to S_3$ as well, by taking $\operatorname{Inj}(3,n)$ instead of $\operatorname{Inj}(2,n)$. It somehow only fails because $S_3$ is not abelian.

The internal hom $\mathbf{Hom}_Y(Y,{-}) \colon G\text{-}\mathbf{Set}/Y \to G\text{-}\mathbf{Set}$ preserves limits, so takes the above to the $A^Y$-torsor $\mathbf{Hom}_Y(Y,X)$ of sections to $\pi$ (this is the torsor in Poonen's answer). Here, $A^Y$ is no longer a constant object: it has a natural $G$-action. Such a torsor corresponds to an element $\phi \in H^1(G,A^Y)$ (nonabelian cohomology, although we will now assume $A$ is abelian).

In the case of $S_n \times S_2$ $\style{display: inline-block; transform: rotate(90deg)}{\circlearrowright}$ $\!\operatorname{Inj}(2,n)$, this gives a homomorphism $S_n \to S_2$. A cocycle computation (included in my original answer) identifies it with the map in Speyer's answer.

The internal hom $\mathbf{Hom}_Y(Y,{-}) \colon G\text{-}\mathbf{Set}/Y \to G\text{-}\mathbf{Set}$ preserves limits, so takes the above to the $A^Y$-torsor $\mathbf{Hom}_Y(Y,X)$ of sections to $\pi$ (this is the torsor in Poonen's answer). Here, $A^Y$ is no longer a constant object: it has a natural $G$-action. Such a torsor corresponds to an element $\phi \in H^1(G,A^Y)$ (nonabelian cohomology, although we will now assume $A$ is abelian), i.e. a crossed homomorphism $\phi \colon G \to A^Y$. Given a section $s \colon Y \to X$ of $\pi$, the crossed homomorphism $\phi \colon G \to A^Y$ can be computed via the composition $$\begin{array}{ccccccc}G & \to & X \underset Y\times X & \stackrel\sim\leftarrow & A \times X & \stackrel{\pi_1}\to & A \\ g & \mapsto & (gs(y),s(gy)).\! & & & & \end{array}$$ In other words, $\phi(g)(y)$ is the unique $a \in A$ such that $gs(y) = as(gy)$. Different choices of section give crossed homomorphisms that differ by a principal crossed homomorphism.

In the case of $S_n \times S_2$ $\style{display: inline-block; transform: rotate(90deg)}{\circlearrowright}$ $\!\operatorname{Inj}(2,n)$, this gives a homomorphism $S_n \to S_2$. Given a section $s \colon {T \choose 2} \to T^2 \setminus \Delta_T$, write $s_1, s_2 \colon {T \choose 2} \to T$ for the projections $\pi_i \circ s$. Then the crossed homomorphism $\phi \colon S_n \to \{\pm 1\}^Y$ can be identified with $$\phi(\sigma)(S) = \frac{x_{\sigma(s_1(S))}-x_{\sigma(s_2(S))}}{x_{s_1(\sigma(S))}-x_{s_2(\sigma(S))}}.$$ Taking the product over all $S \in {T \choose 2}$ recovers the formula in Speyer's answer when $s$ is given by $S \mapsto (\min(S),\max(S))$.