vectors with the same mutual angles Let $S = \{v_1,\ldots,v_n\} \subset \mathbb{R}^n$ and let $T = \{w_1,\ldots,w_n\} \subset \mathbb{R}^n$ be such that the angle between $v_i$ and $v_j$ is equal to the angles between $w_i$ and $w_j$. Suppose we know $S$ is linearly independent.


*

*Can we say $T$ is linearly independent?

*Can we say $T$ is the image of $S$ under an orthogonal transformation? (see edit)

*Is $T$ just $S$ rotated by some angle? (see edit)
This isn't homework, I'd just like to know this since it's related to a problem about reflection groups that I'm working on. For what it's worth, I'm thinking of $S$ as a simple system in a root system.
EDIT: OK, clearly questions 2 and 3 are not true, as the comments make clear. However, I think something may be able to be said about them  ... maybe.
 A: Response to edit.


*

*If you additionally assume that $\| v_i \| = \| w_i \|$ for all $i$, then $T$ is linearly independent, and moreover, $T$ is $S$ multiplied by an orthogonal matrix.  (Multiplication by an orthogonal matrix is an isometry.)  To see this, note that $\| v_i \| = \| w_i \|$ for all $i$ along with the angles between vectors being the same implies that $v_i \cdot v_j = w_i \cdot w_j$ for all $i, j$, and apply the proposition below.

*Without this additional assumption, then $T$ is still linearly independent, because we can scale each $w_i$ to have the same norm as $v_i$ and then apply the first bullet point.  But $T$ will not necessarily be a orthogonal transformation of $S$.
In summary, the answer to question (1) is yes, and the answer to (2) is yes up to scalar multiplication of each vector.

Proposition. Let $S = \{v_1, \ldots, v_n\}$ and $T = \{w_1, \ldots, w_n\}$, such that $v_i \cdot v_j = w_i \cdot w_j$ for all $i, j$ not necessarily distinct.  Suppose further that $S$ is linearly independent.  Then $T$ is linearly independent, and there is an orthogonal matrix $Q$ such that $w_i = Qv_i$ for all $i$.
Proof. Let
$$
A =
\begin{bmatrix}
\quad v_1 \quad \\
\quad v_2 \quad \\
\cdots \\
\quad v_n \quad \\
\end{bmatrix}
\;, \quad 
B =
\begin{bmatrix}
\quad w_1 \quad \\
\quad w_2 \quad \\
\cdots \\
\quad w_n \quad \\
\end{bmatrix}
$$
Since, $v_i \cdot v_j = w_i \cdot w_j$ for all $i, j$, we have
$$
A A^T = B B^T
$$
Since $S$ is linearly independent, $\det A \ne 0$, i.e. $A$ is invertible.
The above implies $(\det A)^2 = (\det B)^2$, so $\det B \ne 0$ also.
We conclude $B$ is invertible,
and we also conclude the first statement that $T$ is linearly independent.
Now consider
$$
Q = B^{-1} A
$$
Note that
\begin{align*}
Q Q^T
&= B^{-1} A A^T (B^{-1})^T
= B^{-1} B B^T (B^T)^{-1} = (I)(I) = I \\
\end{align*}
so $Q$ is orthogonal.
Moreover,
$$
Q(A^T) = B^{-1} A A^T = B^{-1} B B^T = B^T
$$
i.e. $Qv_i = w_i$ for all $i$.
A: If you assume, in addition, that $||v_i||=||w_i||$ for all $i$ then you can derive that $W$ is linearly independent in another way to @6005's answer: if $W$ is linearly dependent then there is some combination $\sum_{i=1}^n a_i w_i=0$ where not all $a_i$ are zero. 
$$ 
0=
(\sum_{i=1}^n a_i w_i,\sum_{i=1}^n a_i w_i) = 
\sum_{i=1}^n\sum_{j=1}^n a_i a_j(w_i,w_j) = 
\sum_{i=1}^n\sum_{j=1}^n a_i a_j(v_i,v_j) = 
(\sum_{i=1}^n a_i v_i,\sum_{i=1}^n a_i v_i) 
$$
Thus $\sum_{i=1}^n a_i v_i=0$ where not all $a_i=0$ in contradiction to $S$ being linearly independent.
This approach has the advantage that $|S|$ and $|W|$ do not have to be the dimension of the space.
