Let $\gamma\_1,\gamma\_2:\mathbb{R}\rightarrow\mathbb{R}^n$ ($n\geq 1$) be two smooth curves such that for every $\,t\_2,t\_1\in\mathbb{R}$ we have $|\gamma\_1(t\_2)-\gamma\_1(t\_1)|=|\gamma\_2(t\_2)-\gamma\_2(t\_1)|$. In otherwords, the pseudometrics on $\mathbb{R}$ given by: $(t\_1,t\_2)\mapsto |\gamma\_1(t\_2)-\gamma\_1(t\_1)|$ and $(t\_1,t\_2)\mapsto |\gamma\_2(t\_2)-\gamma\_2(t\_1)|$ are identical.


**Question:** Must there exist an isometry $f:\mathbb{R^n}\rightarrow \mathbb{R}^n$ such that $\gamma\_2=f\circ\gamma\_1$ ?


Thank you



If the pseudometrics inherited by two smooth curves are identical, must the curves be isometric?
Yes. 


*Sketch of proof*: 


Without loss of generality, we can assume that the affine hull of $\gamma\_1(\mathbb{R})$ has dimension $n$, and that the affine hull of $\gamma\_2(\mathbb{R})$ has dimension $\leq n$. 


Choose $t\_0, \ldots, t\_n$ such that 
$$ \{\gamma\_1(t\_j) - \gamma\_1(t\_0)\}\_{j = 1, \ldots, n} $$
are linearly independent. Let $A$ be the (unique) affine transformation such that $A(\gamma\_1(t\_j)) = \gamma\_2(t\_j)$ for every $j = 0, \ldots, n$. Since it is an affine transformation that fixes the pairwise distance between $n+1$ points, it is an isometry. 


Now, observe that the system of equations 
$$ |x - \gamma\_1(t\_j)| = a\_j , \qquad j\in 0, \ldots, n $$
has at most $1$ solution (this is the simultaneous intersection of $n+1$ spheres) this shows that $A$ must also in fact carry $\gamma\_1(t)$ to $\gamma\_2(t)$ for every $t$. 


