Here is a geometric proof of the result. $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\be}{\boldsymbol{e}}$ As I comment at the end of the proof, this actually proves a stronger fact.

Define two equivalence relations"$\sim_n$" and "$\approx_n$" on the set of bases of $\bR^n$.

$$ (v_1,\dotsc, v_n)\sim_n (w_1,\dotsc, w_n) $$

$\DeclareMathOperator{\vol}{vol}$ $\newcommand{\eps}{\epsilon}$ if for any subset $I\subset \{1,\dotsc, n\}$ we have
$$
\vol(v_i, \;\;i\in I)=\vol(w_i,\;\;i\in I)
$$
and
$$
(v_1,\dotsc, v_n)\approx_n (w_1, \dotsc, w_n) $$
if there exists orthogonal map $T$ and scalars $\eps_i=\pm 1$ such that

$$
w_i= \eps_iTv_i,\;\;\forall i=1,\dotsc , n.
$$
$\newcommand{\Llra}{\Longleftrightarrow}$ The result states that $\sim_n \Llra \approx_n$. Clearly $\approx_n\implies \sim_n$.

Clearly $\sim_1\Llra \approx_1$. We then argue by induction. Assume $\sim_n\Llra \approx_n$. To prove that $\sim_{n+1}\Llra \approx_{n+1}$ it suffices to show that if

$$(v_0, v_1,\dotsc, v_n)\sim _{n+1} (w_0, v_1,\dotsc, v_n), $$

then

$$(v_0, v_1,\dotsc, v_n)\approx_{n+1} (w_0, v_1,\dotsc, v_n). $$

$\DeclareMathOperator{\span}{span}$ $\DeclareMathOperator{\Proj}{Proj}$ For $I\subset \{1,\dotsc, n\}$ we set

$$ V_I:=\span\{v_i;\;\;i\in I\}. $$

and for any vector $u$ we denote by $[u]_I$ its orthogonal projection on $V_I$.

Since $vol(v_0, v_i, i\in I)=\vol(w_0, v_i, I\in I)$, $\forall I\subset \{1,\dotsc, n\}$ we deduce

$$\big\Vert\; v_0-[v_0]_I\;\big\Vert =\Big\Vert\; w_0-[w_0]_I\;\big\Vert. $$
Pythagoras' Theorem implies

$$
\big\Vert\;[v_0]_I\;\big\Vert =\big\Vert\;[w_0]_I\;\big\Vert.
\tag{$\ast$}
$$

Since
$$
\vol(v_0,v_i)=\vol(w_0,v_i),\;\;\forall i=1,\dotsc, n,
$$
there exists $\eps=\pm 1$ such that

$$
(v_0,v_i)=\eps(w_0,v_i),
$$
where $(-,-)$ denotes the inner product. We set
$$
S_i:=\big\{\; \eps=\pm 1;\;\; (v_0,v_i)=\eps(w_0,v_i)\;\big\}.
$$

**Remark.** Above, the two inner products are simultaneously nonzero or simultaneously zero. When they are zero $S_i=\{-1,1\}$. Note that $|S_i|=1$ iff $(v_0,v_i)\neq 0$.

**Lemma 1.** If $i\neq j$ and $(v_i,v_j)\neq 0$, then $S_i\cap S_j\neq \emptyset$.

**Proof.** For simplicity assume $i=1, j=2$. If $(v_0,v_1)=(v_0,v_2)=0$ then $(w_0,v_1)=(w_0,v_2)=0$ and the result is obviously true. Suppose that $(v_0,v_1)\neq 0$. Denote by $\be_1,\be_2$ the orthonormal basis of $\span(v_1,v_2)$ obtained from the basis $v_1,v_2$ via Gram-Schmidt. Then

$$
v_1=x\be_1,\;\; v_2=y\be_1+z\be_2
$$
where $x,z>0$ and $y\neq 0$ since $(v_1,v_2)\neq 0$. Write

$$a_1=(v_0,\be_1),\;\;a_2=(v_0,\be_2), $$
$$ b_1=(w_0,\be_1),\;\;b_2=(w_0,\be_2). $$

Then $a_1=\eps_1 b_1=(v_0,v_1)\neq 0$. Since $a_1^2+a_2^2=b_1^2+b_2^2$ we deduce $a_2=\pm b_2$. We want to show that $a_2=\eps_1 b_2$. Let $a_2=\eta_2b_2$, $\eta_2=\pm 1$. We want to prove that $\eps_1\in S_2$. We argue by contradiction. We have

$$
(v_0,v_2)=\eps_2(w_0,v_2),\;\;\eps_2\neq \eps_1,
$$
so that, given that $a_1=\eps_1b_2$, we get

$$
a_1y+a_2 z=\eps_2\eps_1a_1 y+\eps_2\eta_2 a_2z
$$
Since $\eps_2\neq \eps_1$ we have $\eps_2\eps_1=-1$ and we deduce

$$2 a_1y= (\eps_2\eta_2-1)a_2z\neq 0 $$

so $a_2\neq 0$ and $\eta_2\eps_2=-1$, i.e., $\eta_2=\eps_1 $ and $b_2=\eps_1 a_2$ and $(w_0,v_2)=\eps_1(v_0,v_2)$. $\Box$

We now define a graph $\Gamma$ with vertices $\{1,\dotsc, n\}$ where two distinct vertices $i,j$ are connected by an edge if $(v_i,v_j)\neq 0$. Denote by $\Gamma_1,\dotsc, \Gamma_c$ its connected components. Let $I_\alpha$ denote the set of vertices of $\Gamma_\alpha$. We obtain an orthogonal decomposition

$$ V:=\span(v_1,\dotsc, v_n)=\bigoplus_{\alpha} V_{I_\alpha}. $$

For any vector $u$ we set
$$
[u]_\alpha:=[u]_{I_\alpha}.
$$

**Lemma 2.** Suppose that $i,j$ belong to the same component $\Gamma_\alpha$ and $|S_i|=1$. Then $S_i\subset S_j$.

**Proof.** $\DeclareMathOperator{\dist}{dist}$ We argue by induction on the distance $\dist_{\Gamma_\alpha}(i,j)$. The case $\dist_{\Gamma_\alpha}(i,j)=1$ is covered by **Lemma 1**.

We assume the result is true whenever $\dist(i,j)<d$ and we prove that it is true when $\dist(i,j)=d$.

For simplicity we assume that $i=1$ and $v_1,v_2,\dotsc, v_{d+1}=v_j$ is a path of length $d$ in $\Gamma_\alpha$ connecting $i$ to $j$.

If there exists $k$, $1<k<d+1$ such that $(v_0,v_k)\neq 0$ then $|S_k|=1$, $\dist(v_i,v_k), \dist(v_k,v_j)<d$ and the induction assumption implies

$$ S_i\subset S_k\subset S_j.$$

Since $v_1,\dotsc, v_{d+1}$ is a minimal path connecting $i$ to $j$ we deduce that for any $1\leq k<\ell \leq d+1$, $\ell-k\geq 2$, the vertices $v_j,v_\ell$ are not adjacent so $(v_k, v_\ell)=0$.

Consider the orthonormal basis $\be_1,\dotsc,\be_{d+1}$ of $\span\{v_1,\dotsc, v_{d+1}\}$ obtained from $v_1,\dotsc, v_{d+1}$ via Gram-Schmidt. Then

$$v_1=c_{01}\be_1,\;\;v_2=c_{12}\be_1+c_{02}\be_2,\;\;v_k=c_{1k}\be_{k-1}+c_{0k}\be_k,\;\;k=2,\dotsc, d+1, $$

$$c_{0k}>0,\;\;c_{1k}\neq 0. $$

We denote by $v_0'$ and $w_0'$ the orthogonal projections of $v_0$ and respectively $w_0$ on $\span\{v_1,\dotsc, v_{d+1}\}$.

Then

$$v_0'=\sum_{k=1}^{d+1} a_k\be_k,\;\; w_0'=\sum_{k=1}^{d+1} b_k\be_k. $$

From ($\ast$) we deduce $a_k=\pm b_k$, $\forall k$. Moreover, $a_1=(v_0,v_1)\neq 0$. For simplicity we assume that $a_1=b_i$, i.e. $S_i=S_1=\{1\}$. We have to prove that $1\in S_{d+1}$ i.e.,

$$(v_0, v_{d+1})= (w_0, v_{d+1}). $$

For $k=2,\dotsc, d$ we have

$$ a_{k-1}c{1k}+a_k c_{0k}=(v_0,v_k)=0=((w_0,v_k)=b_{k-1}c_{1k}+b_kc_{0k} $$

and we deduce that

$$a_k=b_k\neq 0,\;\;\forall k=1,\dotsc, d. $$

If $1\in S_{d+1}$ we are done. If $-1\in S_{d+1}$, then

$$ a_dc_{1,d+1}+a_{d+1} c_{0,d+1}=(v_0,v_{d+1})=-(w_0, v_{d+1})=-a_dc_{1,d+1}-b_{d+1} c_{0,d+1}, $$

so that

$$0\neq 2a_dc_{1, d+1}=-(a_{d+1} +b_{d+1}) c_{0,d+1}. $$

Hence $a_{d+1}+b_{d+1}\neq0$ so $a_{d+1}=b_{d+1}$ and thus

$$(v_0, v_{d+1})=(w_0, v_{d+1}), $$

i.e., $1\in S_{d+1}$. $\Box$

**Corollary 3.** For any $\alpha=1,\dotsc, c$ there exists $\eps_\alpha=\pm 1$ such that

$$[v_0]_\alpha=\eps_\alpha[w_0]_\alpha. $$

**Proof.** If $v_0\perp v_i$, $\forall i\in I_\alpha$ the result is obvious since in this case $[v_0]_\alpha=[w_0]_\alpha=0$.

Suppose that $(v_0,v_i)\neq 0$ so that $S_i=\{\eps_i\}$, $\eps_i=\pm 1$. Using **Lemma 2** we deduce that $S_i\subset S_j$, $\forall j\in I_\alpha$.

We can take $\eps_\alpha=\eps_i$. $\Box$

Choose a unit vector $\be_0\in\bR^{n+1}$ such that $\be_0\perp V=\span\{v_1,\dotsc, v_n)$. Then we have an orthogonal decomposition

$$ v_0=a_0\be_0+\sum_\alpha[v_0]_\alpha,\;\;w_0=b_0\be_0+\sum_\alpha [w_0]_\alpha. $$

There exists $\eps_0=\pm 1$ such that $a_0=\eps_0b_0$. Define the orthogonal map $T:\bR^{n+1}\to\bR^{n+1}$ $\newcommand{\bone}{\boldsymbol{1}}$

$$T=\eps_0\bone_{\span\be_0}\oplus \bigoplus_\alpha\eps_\alpha\bone_{V_{I_\alpha}}. $$

Then $Tv_0=w_0$ and $Tv_i=\eps_\alpha v_i$ for $i\in V_\alpha$. $\Box$

**Remark.** The connected components $\Gamma_\alpha$ used in the above proof have a nice geometric interpretation.

For any $I\subset \{1,\dotsc, n\}$ we denote by $G_I$ the group of orthogonal transformations $T$of $V_I$ such that $Tv_i=\pm v_i$, $\forall i\in I$. Clearly $\pm \bone_{V_I}\subset G_I$. The set $I$ is called *irreducible* if $G_I=\{\pm \bone_{V_I}\}$.

Note that if two irreducible sets $I,J$ are not disjoint, then their union is also irreducible. Thus, every $i=1,\dotsc, n$ is contained in a unique maximal irreducible subset and we obtain a partition of $\{1,\dotsc, n\}$ into maximal irreducible sets. These maximal irreducible sets are precisely the vertex sets of the components $\Gamma_\alpha$.

**Comment.** It seems to me that requiring the knowledges of the volumes of all faces is too strong a condition. There are $2^n-1$ faces whereas a basis is determined by $n^2$ numbers. Note that the set of bases of $\bR^n$ modulo the action of $O(n)$ is a space of dimension

$$ n^2-\frac{ n(n-1)}{2}=\frac{n(n+1)}{2}=n+{n \choose 2}. $$

This suggests that, the result ought to be true under weaker assumptions.

andall the diagonals. Since two triangles are congruent if they have the same lengths of sides, we should be able to get a proof from this observation. Or is my sense of rigidity betraying me in higher dimensions? $\endgroup$ – Andrej Bauer May 13 '18 at 8:40