Show that for an isometry $T:{\mathbb{R}}^n \rightarrow {\mathbb{R}}^n$, if
$ X \mapsto \mathrm{dist}(X,T(X)) \quad (X \in {\mathbb{R}}^n)$
is a constant map, $T$ is a parallel translation.
Show that for an isometry $T:{\mathbb{R}}^n \rightarrow {\mathbb{R}}^n$, if
$ X \mapsto \mathrm{dist}(X,T(X)) \quad (X \in {\mathbb{R}}^n)$
is a constant map, $T$ is a parallel translation.
Isometries of $R^n$ are known to be linear functions of the form $T(x)=A(x)+b$ where $A$ is a linear isometry fixing 0 and $b$ is a vector. For an isometry $T$ the displacement $f(x) = T(x) - x$ is a linear vector-valued function of $x$. If $f$ is of constant length then $f$ must be constant, otherwise there is a pair of points $p,q$ with $f(p) \neq f(q)$ and then $f$ would be unbounded on the line through $p$ and $q$.