Minimal specification of isometry in terms of norm preservation Let $V,W$ be $n$-dimensional (real) inner product spaces, and let $T:V \to W$ be a linear map. 
Let $v_1,...,v_n$ be a basis of $V$. It is easy to see that if $|T(v)|_W=|v|_V$ for every $v \in \{v_1,...,v_n,v_1+v_2,v_1+v_3,...,v_{n-1}+v_n\}$, then $T$ is an isometry (a proof is provided below). 
In other words, after choosing wisely $k(n):=\frac{n(n+1)}{2}$ vectors, it is enough to verify $T$ preserves the norms of these special vectors, in order to conclude it's an isometry.
Question: Is there no way to choose less than $k(n)$ vectors, in such a way that every linear map which preserves their norms is an isometry?

I believe we cannot choose less vectors. I have some "convincing evidence" for the cases $n=1,2,3$ (see below), but I am not sure how to give a rigorous argument.
Note that a "wise choice" of vectors does not have to be of the form of some vectors, and linear combinations of them (I do feel this it the most efficient method, but I don't see how to prove this). Even if we prove that this is the case, than we need to show we cannot do better than to work with only orthonormal bases.

The partial "evidence":
$n=1: k=1$. Obvious
$n=2: k=3$. Take $V=W=\mathbb{R}^n$ with its standard inner product. Then, $T(e_1)=e_1, T(e_2)=\frac{e_1+e_2}{\sqrt 2}$ is a counter example.
$n=3: k=6$. Then any matrix of the form $$ \begin{pmatrix} c & s & x \\  -s & c & y  \\  0 & 0 & z \\\end{pmatrix} $$ where $c^2+s^2=1,x^2+y^2+z^2=1, sx+cy=0$ preserves the norms $e_1,e_2,e_1+e_2,e_3,e_2+e_3$ but it's an isometry only if $|z|=1,x=y=0$.

Proof that $k(n)=\frac{n(n+1)}{2}$ vectors are enough:
Noting that $$ \langle u,v \rangle = \frac{1}{2}(|u+v|^2 - |u|^2 - |v|^2) ,$$ we obtain
$$ \langle Tv_i,Tv_j \rangle = \frac{1}{2}(|Tv_i+Tv_j|^2 - |Tv_i|^2 - |Tv_j|^2) = \frac{1}{2}(|T(v_i+v_j)|^2 - |v_i|^2 - |v_j|^2)  $$
$$ = \frac{1}{2}(|v_i+v_j|^2 - |v_i|^2 - |v_j|^2) = \langle v_i,v_j \rangle,$$
thus $T$ is an isometry.
 A: Let $f:\mathbb R^n\to V$ and $g:W\to\mathbb R^n$ be any two linear isometries. Then $T$ is an isometry if and only if $g\circ T\circ f$ is an isometry. So, we may assume without loss of generality that $V=W=\mathbb R^n$ equipped with the Euclidean inner product.
Let $v_1,\ldots,v_k\in\mathbb R^n$ be $k<\frac12n(n+1)$ arbitrarily chosen vectors. It suffices to exhibit the existence of a non-isometric linear transformation $T$ on $\mathbb R^n$ that preserves the norm of each $v_j$. Consider the system of homogeneous linear equations
$$
v_j^\top Av_j=0,\quad j=1,\ldots,k,\tag{1}
$$
where the $n^2$ entries of $A\in M_n(\mathbb R)$ are unknown. Since the subspace of all $n\times n$ skew-symmetric matrices has dimension $\frac12n(n-1)<n^2-k$, the system $(1)$ must admit a nontrivial solution $A$ that is not skew-symmetric. However, if $A$ is a solution, so is $A+A^\top$. Therefore, $(1)$ admits a nontrivial symmetric solution $A$.
Let $P=I+\varepsilon A$, where $\varepsilon>0$ is sufficiently small. Then $P$ is positive definite. Define $Tx=\sqrt{P}x$. Then $T$ is not an isometry because $\sqrt{P}$ is not real orthogonal. However, for each $j$ we have
$$
\|Tv_j\|^2=v_j^\top Pv_j=v_j^\top(I+\varepsilon A)v_j=v_j^\top v_j=\|v_j\|^2.
$$
A: I'll try to elaborate on my idea in the comment. Preserving the scalar product means that the two symmetric bilinear forms must coincide: $$(x,y) \mapsto \langle x,y \rangle$$ $$(x,y) \mapsto \langle Tx,Ty \rangle$$
This is equivalent to saying that their matrices must coincide. But a symmetric matrix has $\frac{n(n+1)}{2}$ degrees of freedom.
This means, once we've chosen the basis in $V$ we must choose the products $\langle v_i,v_j \rangle$ for $i \geq j$.


*

*Is it enough? Yes, because we've chosen the coefficients of the matrix relative to its canonic basis (here I'm taking about the basis of matrices which have $1$ in one component and $0$ in the others, but only for $i \geq j$ since the matrix is symmetric).

*Can we take less? No, because the symmetric matrices space has dimension $\frac{n(n+1)}{2}$ as we've said.


How does this relate to our question? Choosing a basis for $V$ and then such products is equivalent to choosing the list of vectors you proposed (along with their norms).
(I.e. Choosing a bilinear form is equivalent to choosing the quadratic forum associated to it, which in the case of a scalar product is the norm)
