This is a sketch of a quasi-geometrical proof. Later today I'll fill in the details. After finding it I've realized that it is somewhat similar to the combinatorial-computational proof in the article that I've mentioned. The key is the following lemma:

**Lemma.** Assume that $v_1,...,v_n$ are linearly independent in $\mathbb{R}^{n+1}$ and $u,w\in \mathbb{R}^{n+1}$ are such that for any $A=\{i_1,...,i_k\}\subset\{1,...,n\}$ we have $u_A\bot w_A$, where $u_A$ and  $w_A$ are the projection of $u$ and $v$ respectively on $\mathrm{span}\{v_i,|i\in A\}$. Let $\{i_1,...,i_k\}\sqcup \{j_1,...,j_l\}\subset \{1,...,n\}$ be such that:

 - $v_{i_p}\bot v_{i_q}$ whenever $|p-q|>1$ and $v_{i_p}\not\bot v_{i_{p+1}}$;
 - $v_{j_p}\bot v_{j_q}$ whenever $|p-q|>1$ and $v_{j_p}\not\bot v_{j_{p+1}}$;
 - $v_{i_p}\bot u$ whenever $p>1$ and $v_{i_1}\not\bot u$;
 - $v_{i_p}\bot w$ whenever $p>1$ and $v_{j_1}\not\bot w$.

Then $\mathrm{span}\{u,v_{i_1},...,v_{i_k}\}\bot~\mathrm{span}\{w,v_{j_1},...,v_{j_k}\}$.
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After having this lemma let's argue by induction: it is enough to show that if $v_1,...,v_n$ are linearly independent in $\mathbb{R}^{n+1}$ and $v_0,v_{n+1}\in \mathbb{R}^{n+1}$ are such that for any $\{i_1,...,i_k\}\subset\{1,...,n\}$ the $k+1$-dimensional volumes of $P(v_{n+1},v_{i_1},...,v_{i_k})$ and $P(v_0,v_{i_1},...,v_{i_k})$ is the same, then there $a_0,a_1,...,a_n=\pm 1$ and an orthogonal operator $U$ on $\mathbb{R}^{n+1},$ such that $v_i=a_iUv_i$, for every $i\in\overline{1,n}$ and $v_{n+1}=a_0Uv_0$.

Let $2u=v_0+v_{n+1}$ and $2w=v_0-v_{n+1}$. Then $u,w$ satisfy the conditions of the Lemma, and so we can divide $v_1,...,v_n$ into three collections, say $v_1,...,v_k$, $v_{k+1},...,v_l$ and $v_{l+1},...,v_n$ such that $\mathrm{span}\{u,v_1,...,v_k\}\bot~\mathrm{span}\{w,v_{l+1},...,v_{n}\}$, $\mathrm{span}\{u,v_1,...,v_k\}\bot~\mathrm{span}\{v_{k+1},...,v_{l}\}$ and $\mathrm{span}\{v_{k+1},...,v_l\}\bot~\mathrm{span}\{w,v_{l+1},...,v_{n}\}$.
Now define $U$ by $Uv_1=v_1,...,Uv_l=v_l$, $Uv_{l+1}=-v_{l+1},...,Uv_n=-v_n$, and $Uu'=w'$, where $u',w'$ are projections of $u,w$ on ${v_1,...,v_n\}^{\bot}$. This $U$ is the operator we need.

Like I said in the beginning, I'll add the missing details in several hours.