Motivated by [this paper and its economics motivations](http://www.sciencedirect.com/science/article/pii/S0723086904800161), 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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/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 symetric 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?