Motivated by this paper and its economics motivations, we recall that a social choice among $n$ objects is a continuous function $$f:\overbrace{M\times M\times\cdots\times M}^{\text{$n$ times}}\to M$$
which satisfy the following conditions:
$f(x,x,\ldots,x)=x$
$f$ is unchanged under all permutations $\sigma \in S_{n},$ the symmetric group on $n$ elements.
This is a mathematical modeling of the following economic situation:
A client has to choose one item among $n$ items $(x_{1},x_{2},\ldots,x_{n})$. His preference function is denoted by the above $f$.
Now it is natural that we assume that the client faces with n different items, so $x_{i} \neq x_{j}$, $\forall i \neq j$. So we consider the ordered configuration space $$F_{n}(M)=\{(x_{1},x_{2},\ldots,x_{n})\in M^{n}\mid x_{i} \neq x_{j},\;\forall i \neq j\} $$.
There is an obvious action of the symmetric group on $F_{n}(M)$.
Now the following question can be counted as an equivariant analogy of the social choice problem:
Assume that $M$ is a manifold which is acted by the symmetric group $S_{n}$. Is there always an equivariant continuous map $f:F_{n}(M)\to M$? If not, for what type of manifolds the answer is affirmative? What type of algebraic topological obstructions would appear?
Can we find an economics interpretation for this equivariant version?
Example: For $M=S^{1},\; n=2$, the symmetric group $S_{2}$ acts on $S^{1}$ by complex conjugation $z\mapsto \bar{z}$ Then the function $f(z,w)=z\bar{w}$ satisfies $f(z,w)=\bar{f(w,z)}$.
In the later example let's replace the conjugate action by antipodal action. Is there a continuous map $f:F_{2}(S^{1})\to S^{1}$ which satisfy $f(z,w)=-f(w,z)$?