Suppose that $F:S^{n-1}\to A$ is a map of sets from the unit sphere in $\mathbb R^n$ to an abelian group, and that the sum $F(v\_1)+\dots +F(v\_n)$ over an orthonormal basis is independent of the basis. Does it follow that $F$ is a constant function? 


This is clearly false for $n=2$. I am wondering if it is true for sufficiently large $n$.


ADDED LATER The $\mathbb R$-valued examples in Cranch's answer may be combined into a single example $v\mapsto v\otimes v$ with values in $\mathbb R^n\otimes \mathbb R^n$, or $n\times n$ matrices. Its image generates the group of symmetric matrices with integer trace. It seems reasonable to expect that every continuous real-valued example comes from this one -- in other words has the form $v\mapsto B(v,v)$ for symmetric bilinear $B$. Maybe this can be worked out using Sawin's suggestion about representations of $O(n)$. But I was also curious about the general case, where the target group might not be (uniquely) divisible.



A combinatorial question about orthonormal bases
Given a vector $u$ and an orthonormal basis $x\_1,\ldots,x\_n$, we have $||u||^2 = \left<u,x\_1\right>^2 + \cdots + \left<u,x\_n\right>^2$. But that means that, if you choose a nonzero $u$, then the function $F(x) = \left<u,x\right>^2$ gives a counterexample.


