Can any one prove this claim? Let $S^n\subset \mathbb{R}^{n+1}$ be the unit sphere. For any given point
$$x=(x_1, \cdots, x_i, \cdots, x_j, \cdots, x_{n+1}) \in S^n,$$
exchange any two coordinate $x_i$ and $x_j$  of $x$ to get a new point
$$x'= (x_1, \cdots, x_j, \cdots, x_i, \cdots, x_{n+1})$$
(may or may not be the same as $x$).
For a given $x\in S^n$, we denote the set of points (including $x$ itself) obtained by this way as $\Gamma_x$.
My claim is: if $x\in S^n$ satisfies $\sum_1^{n+1} x_i = 0$, $\Gamma_x$ can linearly span to the hyperplane $\sum_1^{n+1} x_i =0$.
How to prove this claim?
 A: The condition that $x$ lies on the sphere can be replaced with the condition that $x$ is nonzero. For if $x$ is nonzero, one can divide $x$ by its length. The condition that $x$ is nonzero and that the $x_i$ sum to $0$ implies that the $x_i$ take at least two distinct values.
For $i\ne j$ define $y_{ij}$ to be the vector with $x_i$ and $x_j$ swapped. Denote the standard basis vector for $1\le i\le n+1$ by $e_i$ and write
$$
z_{ij}=y_{ij}-x=(x_j-x_i)e_i+(x_i-x_j)e_j.
$$
If $x_i$ is distinct from all the $x_j$, $j\ne i$, then the set of $n$ vectors $z_{ij}$, $j\ne i$, spans the hyperplane $\sum_{i=1}^{n+1}x_i=0$.
If an $x_i$ distinct from $x_j$ for all $j\ne i$ cannot be found, then let $I$ be the set of indices $i$ for which $x_i=x_1$ and let $J$ be the set of indices $j$ for which $x_j\ne x_1$. Let $j_1$ be an element of $J$. Then the set of $n$ vectors
$$
\{z_{1j}\mid j\in J\}\cup\{z_{ij_1}\mid i\in I\setminus\{1\}\}
$$
spans the hyperplane $\sum_{i=1}^{n+1}x_i=0$.
For example, if $x=\begin{bmatrix}1 & 1 & 1 & -1 & -1 & -1\end{bmatrix}$, a set of spanning vectors is $\{z_{14},z_{15},z_{16},z_{24},z_{34}\}$ with
\begin{align}
z_{14}&=\begin{bmatrix}-2 & 0 & 0 & 2 & 0 & 0\end{bmatrix}\\
z_{15}&=\begin{bmatrix}-2 & 0 & 0 & 0 & 2 & 0\end{bmatrix}\\
z_{16}&=\begin{bmatrix}-2 & 0 & 0 & 0 & 0 & 2\end{bmatrix}\\
z_{24}&=\begin{bmatrix}0 & -2 & 0 & 2 & 0 & 0\end{bmatrix}\\
z_{34}&=\begin{bmatrix}0 & 0 & -2 & 2 & 0 & 0\end{bmatrix}.
\end{align}
