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Let $A,B$ be two convex compact subsets with non-empty interior in a Euclidean space $\mathbb{R}^n$. Let $f\colon A\to B$ be a bijective isometry between them.

Does there exist an isometry $F\colon \mathbb{R}^n\to \mathbb{R}^n$ such that its restriction to $A$ is equal to $f$?

Remark. Necessarily $F$ is a composition of translation and orthogonal transformation.

A reference would be helpful.

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    $\begingroup$ convexity is not needed: non-empty interior is enough, moreover, having $n+1$ linearly independent points is enough. distances to these points determine the point in a space uniquely, so knowing images of these points you determine the images of every point of space. $\endgroup$
    – Lev Soukhanov
    Commented Jan 24, 2020 at 8:03
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    $\begingroup$ @LevSoukhanov: Thanks, you are right. This is a final answer in fact. $\endgroup$
    – asv
    Commented Jan 24, 2020 at 8:12
  • $\begingroup$ Even without $n+1$ linearly independent points, there exists an isometry such that its restriction is $f$; but in that case it is not unique. $\endgroup$
    – Gerald Edgar
    Commented Jan 24, 2020 at 11:53

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