Existence of some special kind of isometry in R^n-1

[julypraise]

Homework Statement

Is there a theorem that states the following?

Let $P= \{ P_{1}, . . . , P_{n} \}$ be the set of n distinct points in $\mathbb{R}^{n-1}$ and $P'= \{ P'_{1}, . . . , P'_{n} \}$ also a set of points in $\mathbb{R}^{n-1}$. If for all $i,j$ $|P_{i} - P_{j}|=|P'_{i} - P'_{j}|$ then there is an isometry $f: \mathbb{R}^{n-1} \to \mathbb{R}^{n-1}$ such that for all $i$ $f(P_{i})=P'_{i}$.

Or at least is it ture? If true how to prove it?

Homework Equations

The Attempt at a Solution

Geometrically, it seems plausible especially in some cases where such an isometry is a rotation, a translation, a reflection, or a combination of them.

 

[Nino MMP]

To prove it rigorously, I think one would have to define an isometry f: \mathbb{R}^{n-1} \to \mathbb{R}^{n-1} and then show that for all i, f(P_{i})=P'_{i} and also show that |f(P_{i}) - f(P_{j})|=|P'_{i} - P'_{j}|.