Independent vectors in the permuting coordinates action of $S_n$ on $\mathbb{R}^n$ Let $V$ be the hyperplane in $\mathbb{R}^n$ with equation $\sum_i x_i=0$. The symmetric group $S_n$ acts on $V$ by $s\cdot (v_1,\ldots,v_n)=(v_{s^{-1}(1)},\ldots,v_{s^{-1}(n)})$. Consider those $v\in V$ with the following property:

For any $n-1$ elements $s_1=\text{id},s_2,\ldots,s_{n-1}\in S_n$, the vectors
$v,s_2\cdot v,\ldots,s_{n-1}\cdot v$ are independent.

Question: what $v$ have this property?
Some simple observations:

*

*If $H_{ij}$, for $1\leq i<j\leq n$, is the hyperplane in $\mathbb{R}^n$ with equation $x_i-x_j=0$ (i.e. a reflecting hyperplane of $S_n$ in the permuting coordinates action on $\mathbb{R}^n$) then a $v$ with the property above cannot lie in $V\cap H_{ij}$.

*If $X$ is a proper non-empty subset of $\{1,\ldots,n\}$ and $H_X$ is the hyperplane with equation $\sum_{i\in X}x_i=0$ then a $v$ with the property above cannot lie in $V\cap H_X$.

*When $n=3$ the $v\in V$ that remain after removing those in (1) and (2) satisfy the property above.

 A: This is not an answer, but I think comments can't include images.  I have marked it as CW to avoid reputation for a non-answer.  (Actually I think @SeanEberhard's emendations below, describing the forbidden surfaces algebraically, now makes this probably as good an answer as is likely to arise.)
Per your answer for $n = 3$ and @SeanEberhard's comment, we need only consider $n = 4$.
In the ‘obvious’ coördinates $(a, b, c)$ on $V$ (i.e., writing a typical element of $V$ as $(a, b, c, -(a + b + c))$), here is the picture, generated by extremely naïve Mathematica code, of the region that doesn't satisfy your condition.

Region[ImplicitRegion[
  Or @@
   (# == 0 & /@
     (Det /@ ((Most /@ # &) /@
         (Prepend[#, {a, b, c, -(a + b + c)}] & /@
           Select[
            Tuples[
             Select[Permutations[{a, b, c, -(a + b + c)}], 
              Not[# === {a, b, c, -(a + b + c)}] &],
             2],
            Not[#[[1]] === #[[2]]] &])))),
  {a, b, c}]]

Computer says it is the union of the following surfaces, together with all permutations of $(a, b, c, -(a+b+c))$ (corresponding $s_2, s_3$ also listed):
$$
a = 0 \qquad [(3,4), (2,3,4)],\\
b - c = 0 \qquad [(1,3)(2,4), (1,2,4,3)],\\
b + c = 0 \qquad [(1,2), (1,4)(2,3)],\\
b^{2} - a c = 0 \qquad [(3,4), (1,2,3,4)],\\
a^{2} - a b + b^{2} - a c - b c + c^{2} = 0 \qquad [(1,3,2), (1,2,3)],\\
a^{3} + a^{2} b + a b^{2} + b^{3} - a b c - a c^{2} - b c^{2} - c^{3} = 0 \qquad [(1,4,3,2), (1,2,3)],\\
a^{2} + a b + b^{2} + a c + 3 b c + c^{2} = 0 \qquad [(1,3)(2,4), (1,3,4)],\\
2 a^{2} + 2 a b + b^{2} + 2 a c + c^{2} = 0 \qquad [(1,3,4,2), (1,2,4,3)],\\
2 a^{2} b + 2 a b^{2} - a^{2} c + a b c - b^{2} c - a c^{2} - b c^{2} - c^{3} = 0 \qquad [(1,4,3), (1,2,3)].
$$
