An equivariant social choice in Mathematical economics 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:
1) $f(x,x,\ldots,x)=x$
2) $f$ is unchanged  under all permutations  $\sigma \in S_{n},$ the symmetric group on $n$ elements.
Perhaps the following could be  considered as a social model for the above mathematical problem.
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 $S_{n}$ 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 $S_{2}$ can act  on $S^{1}$  by complex conjugation $z\mapsto \bar{z}$ Then the function $f(z,w)=z\bar{w}$ satisfies $f(z,w)=\overline{f(w,z)}$. 
In this example let's replace the conjugate action by antipodal action. Then $f(z,w)= (z-w)/|z-w|$ is  a  map $f:F_{2}(S^{1})\to S^{1}$  which satisfy $f(z,w)=-f(w,z)$?
 A: I'll attempt an answer to the mathematical question, without discussing the motivation. As I understand it, $M$ is a manifold with $S_n$-action, and we are asking whether there exists an $S_n$-equivariant map $f:F_n(M)\to M$, where the action on $F_n(M)$ is by permutation of coordinates (and in particular has nothing to do with the action on $M$).
As a first observation, note that if $x\in M$ is a fixed point $x$, then such an $f$ exists; just map everything to $x$.
A general necessary condition for the existence of a $G$-map $f:X\to Y$ between $G$-spaces is given by the Faddell-Husseini index, as decribed for example in this paper:

Pavle V. M. Blagojević, Wolfgang Lück, Günter M. Ziegler, Equivariant Topology of Configuration Spaces, J. Topology 8 (2015) pp 377–413, doi:10.1112/jtopol/jtu029, arXiv:1207.2852.

Given a $G$-space $X$ and a commutative ring $R$, the Faddell-Husseini index of $X$ is the ideal in $H^*(BG;R)$ defined by
$$
\operatorname{Index}_G(X;R):=\ker(p^*:H^*(BG;R)\to H^*(EG\times_G X;R)),
$$
where $p:EG\times_G X\to BG$ is the projection of the Borel fibration. Then it is easy to see that if there exists a $G$ map $f:X\to Y$ then 
$$
\operatorname{Index}_G(Y;R)\subseteq \operatorname{Index}_G(X;R)
$$
must hold. You can sometimes rule out existence of equivariant maps using this property. The index of the configuration space $F_n(M)$ is probably difficult to compute in general, but is done in the linked paper for $M=\mathbb{R}^d$. 
