Let $V$ be an $n$ dimensional real vector space. Let $S = \{ v_1, ... , v_m \}$ be a spanning set for $V$ not containing the zero vector. Suppose that $\phi_i: V \rightarrow V, i = 1, 2$ are linear transformations which send $\alpha := v_1$ to $-\alpha = -v_1$, which fix pointwise a subspace $W_i$ of dimension $n-1$, and which map $\{v_1, ... , v_m\}$ into itself. Why is it the case that $\phi_1 = \phi_2$?
Attempt: Let $T = \{v_t, ... , v_m\}$ be a maximal subset of $S$ whose span $W'$ does not contain $\alpha$. Then $W'$ is $n-1$ dimensional, and we have a linear transformation $\phi': V \rightarrow V$ which fixes $W'$ pointwise and sends $\alpha$ to $-\alpha$.
So we just have to show that if $\phi$ is a linear operator sending $\alpha$ to $-\alpha$, fixing pointwise a subspace $W$ of dimension $n-1$, and permuting $S$, then $\phi = \phi'$.
The only thing I can think of is to suppose that $W' \neq W$. Then by a dimension argument, $V = W + W'$. We can write $$\alpha = w + \sum\limits_{i=t}^m c_i \alpha_i$$ so $-\alpha = w + \sum\limits_i c_i\alpha_{\sigma i}$, where $\sigma$ is the permutation of $S$ given by $\phi$. Adding these equations, we get that $w$ is in the span of $\alpha_i, \alpha_{\sigma i}, i = t, ... , m$.
Let $G$ be the subgroup of $GL(V)$ generated by $\phi_1$, $\phi_2$. $G$ is finite, as its action on $S$ defines an injection $G \to Aut(S)$ into a the finite group of permutations of $S$. Hence there is a $G$-invariant scalar product $\langle-,-\rangle \colon V \times V \to \mathbb{R}$. Thus $\phi_1$ and $\phi_2$ must fix the orthogonal complement of $v_1$ with respect to $\langle-,-\rangle$, thus they are equal.