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 $\mathrm{span}\{u,v_{i_1},...,v_{i_k}\}\bot~\mathrm{span}\{w,v_{j_1},...,v_{j_k}\}$.

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.

This is a quasi-geometrical proof. After finding it I've realized that it is somewhat similar to the combinatorial-computational proof in the article that I've mentioned. Let $H$ be a real inner product space. We will call a finite sequence $u_1,...,u_n\in H$ a chain (from $u_1$ to $u_n$) if $u_{i}\bot u_{j}$ whenever $|i-j|>1$ and $u_i\not\bot u_{i+1}$. We need two lemmas:

Lemma 1. Let $u_1,...,u_n\in H$ be a chain. Then $u_1,...,u_{n-1}$ are linearly independent.

Proof. We will show that $u_k\not\in \mathrm{span}\{u_1,...,u_{k-1}\}$ for every $1<k<n$. Assume that $u_k=\alpha_1 u_1+...+\alpha_{k-1} u_{k-1}$, where $1<k<n$. Then each of $u_1,u_2,...,u_{k-1}$ are perpendicular to $u_{k+1}$, and so $u_k\bot u_{k+1}$, which contradicts the definition of chain.

Lemma 2. Let $B\subset H$ be linearly independent. Let $u,w\in \mathrm{span}B$ be such that for any $A\subset B$ we have $u_A\bot w_A$, where $u_A$ and $w_A$ are the projection of $u$ and $v$ respectively on $\mathrm{span}A$. Then there is a partition $B=B_u\sqcup B_w$ such that $B_{u}\bot B_{w}$ and $u\in \mathrm{span}B_{u}, w\in\mathrm{span}B_{w}$.

Proof. Define $B_u$ to be the set of all $v\in B$ such that there is a chain from $u$ to $v$ from elements of $B$, and $B_w=B\backslash B_u$.

In order to prove proposition it is enough to show that if $u_1,...,u_n,w_1,...,w_m\in B$ are such that $u_0=u,u_1,...,u_n$ and $w_0=w,w_1,...,w_m$ are chains, then $u_n\bot w_m$. We will use the induction by $m+n$. When $m+n=0$ this follows from $u_0=u=u_B\bot w_B=w=w_0$.

Assume the claim holds for $m+n$ and assume that $A=\{u_1,...,u_n,u_{n+1},w_1,...,w_m\}\subset B$ are such that $u_0=u,u_1,...,u_n,u_{n+1}$ and $w_0=w,w_1,...,w_m$ are chains. Then, from the hypothesis of induction $u_i\bot w_j$, when $i\le n$. Let $u'\bot v'$ be the orthogonal projections of $u,w$ on $\mathrm{span}A$. Then $u'\bot u_i$ for $i>2$ and $u'\bot w_i$. Hence, $u'\not\bot u_1$, and so $u',u_1,...,u_n,u_{n+1}$ is a chain. Analogously, $w',w_1,...,w_m$ is also a chain.

Note that $u',u_1,...,u_n\in \{w',w_1,...,w_m\}^{\bot}$. All these $m+n+2$ vectors belong to span of the linearly independent collection $A$, whose dimension is $m+n+1$. By Lemma 1, $u',u_1,...,u_n$ are linearly independent, as well as $w',w_1,...,w_{m-1}$, and so $w_m\in \mathrm{span}\{w',w_1,...,w_{m-1}\}$. Since all of the vectors in the span are perpendicular to $u_{n+1}$, we conclude that $u_{n+1}\bot w_m$.

After having Lemma 2 let's prove the Proposition 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 $a_1,...,a_n=\pm 1$ and an orthogonal operator $T$ on $\mathbb{R}^{n+1},$ such that $v_i=a_iTv_i$, for every $i\in\overline{1,n}$ and $v_{n+1}=Tv_0$.

Let $v'_0$ and $v'_{n+1}$ be the projections of $v_0$ and $v_{n+1}$ on $\mathrm{span}\{v_1,...,v_n\}$. Note that $\|v_0-v'_0\|=\|v_{n+1}-v'_{n+1}\|$. Also let $2u=v'_0+v'_{n+1}$ and $2w=v'_0-v'_{n+1}$. Then $u,w$ satisfy the conditions of the Lemma. Indeed, the projection of $v'_0=u+v$ and $v'_{n+1}=u-v$ on the span of any combination of $v_i$ have equal length, and $\|Pr~u+Pr~v\|=\|Pr~u-Pr~v\|$ implies $Pr~u\bot Pr~v$.

By Lemma 2 we can find $1\le k\le n$ and relabel $v_1,...,v_n$ so that $v_1,...,v_k\bot v_{k+1},...,v_{n}$ and $u\in \mathrm{span}\{v_1,...,v_k\},~w\in \mathrm{span}\{v_{k+1},...,v_n\}$.

Now define $T$ by:

• $Tv_1=v_1,...,Tv_k=v_k$ (from which it follows that $Tu=u$);
• $Tv_{k+1}=-v_{k+1},...,Tv_n=-v_n$ (from which it follows that $Tw=-w$);
• $T(v_0-v'_0)=v_{n+1}-v'_{n+1}$; then $Tv_0=T(v_0-v'_0)+Tu+Tw=v_{n+1}-v'_{n+1}+u-w=v_{n+1}$.

Since $T$ restricted to $\mathrm{span}\{v_1,...,v_k\}$, $\mathrm{span}\{v_{k+1},...,v_n\}$ and $\{v_1,...,v_n\}^{\bot}$ is orthogonal, and these subspaces are also mutually orthogonal, we see that $T$ is an orthogonal. Thus, $T$ satisfies all the desired properties.