Problem. Let $n\in\mathbb N$, $X$ be a strictly convex $n$-dimensional real Banach space, $S_X=\{x\in X:\|x\|=1\}$ be the unit sphere of $X$, and $e_1,\dots,e_n\in S_X$ be linearly independent points. Let $f:S_X\to S_X$ be an isometry such that $f(e_i)=e_i$ for all $i\le n$. Is $f$ the identity map of $S_X$?
Remark 1. The answer is affirmative if $n\le 3$. For $n=3$ this follows from a known fact that any action of a compact group on the $2$-sphere is equivalent to the linear action of a compact subgroup of $O(3)$ on $S^2$.
Remark 2. A Banach space is strictly convex if every convex subset of its unit sphere contains at most one point.
Example (Javier Cabello Sanchez): Consider the strictly convex 3-dimensional Banach space $X=(\mathbb R^3,\Vert\cdot\Vert)$ endowed with the $\ell_3$-norm $$\Vert(x,y,z)\Vert=\sqrt[3]{|x|^3+|y|^3+|z|^3}$$and basis $$\mathbf e_1=\tfrac1{\sqrt[3]{6}}(1,1,-\sqrt[3]{4}),\quad \mathbf e_2=\tfrac1{\sqrt[3]{6}}(1,-\sqrt[3]{4},1),\quad \mathbf e_3=\tfrac1{\sqrt[3]{6}}(-\sqrt[3]{4},1,1).$$ For the point $\mathbf x=\tfrac1{\sqrt[3]{3}}(1,1,1)\in S_X$ and every $i\in\{1,2,3\}$ we have $$\Vert \mathbf e_i-\mathbf x\Vert=\Vert \mathbf e_i+\mathbf x\Vert=\sqrt[3]{\tfrac43+\sqrt[3]{2}},$$ which means that the point $\mathbf x$ on the sphere $S_X$ is not uniquely determined by its distances to the six points $\mathbf e_1,\mathbf e_2,\mathbf e_3, -\mathbf e_1,-\mathbf e_2,-\mathbf e_3$. This Example shows that the Problem cannot be reduced to the level of individual points.
