Minimal specification of isometry in terms of norm preservation

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 $$ \|Tv_j\|^2=v_j^\top Pv_j=v_j^\top(I+\varepsilon A)v_j=v_j^\top v_j=\|v_j\|^2. $$