Extend isometry on some cube vertices to the entire cube

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Let $K\subset V=\{-1,1\}^n$ be a set of vertices of the $n$-dimensional hypercube $D=[-1,+1]^n$ and let $f:K\to V$ be an isometry with respect to the Euclidean metric inherited from $\mathbb R^n.$ We do not know a priori that $f$ corresponds to a linear endomorphism of $\mathbb R^n.$

Can $f$ necessarily be expressed as a composition of reflections by coordinate hyperplanes and permutations of coordinates, i.e., does there exist a global symmetry of $D$ that restricts to $f$ on $K$?

This question came up here when analysing the distribution of some random vectors.

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Here is a counterexample. Let $n=4$ and let $K$ consist of the rows of the following array: $$\begin{matrix} -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & 1 & -1 \\ 1 & -1 & -1 & 1 \\ \end{matrix}$$

Note that any isometry $D\to D$ which fixes the first $3$ points must be the identity (fixing the first point says you don't reverse the sign of any coordinate, and each coordinate is uniquely determined by the subset of the $3$ points that take the value $1$ on that coordinate, so the permutation of the coordinates must be trivial). But there is an isometry $K\to V$ which fixes the first $3$ points and sends the last point to $$\begin{matrix} -1 & 1 & 1 & -1 \end{matrix}$$