The symmetric group $S_{m}$ is equiped with the counting Har measure $\mu$ and the obvious sgn character. Assume that $S_{m}$ acts on $S^{n}$, $n\geq m-1$ and $f:S^{n}\to \mathbb{R}^{n}$ is a continuous map.

Is it true to say that there exist $x_{0}\in S^{n}$ such that $$ \int_{S_{m}} sgn(g)f(g.x_{0})d\mu=0 $$ In the other word: Is there a point $x_{0}\in S^{n}$ such that $$ \sum_{g\in A_{n}} f(g.x_{0})= \sum_{g\notin A_{n}} f(g.x_{0}) $$ where $A_{n}$ is the group of even permutations?

**Note 1:** In this question the action is arbitrary but continuous, so we dont consider a particular action.

**Motivations:** The answer is affirmative in each of the following cases:

1.$m=2$ (The original Borsuk Ulam theorem or its involutive version)

2.If we replace the statement "$f:S^{n}\to \mathbb{R}^{n}$" with "$f:S^{n}\to R^{2}$".

3.Every action with singularity, for example the obvious action of $S_{m}$ on $S^{m-1}.$

**Note 2:** A natural question related to this post could be "To what extent, the action of permutation groups on spheres is classified"? Of course there is a non singular action of $S_{m}$ on $S^{m-1}$ with $sgn(g)g.x$ where $g.x$ is the obvious action.