Given a finite set of $n$ points on the plane. Is it fully characterized up to translations/rotations/reflections (isometric transformations) by the (multi)set of its $(n-1)n/2$ distances between pairs of distinct points?
Example: If we consider the corners of a square with side 1, the multiset of distances is $\{1,1,1,1,\sqrt{2},\sqrt{2}\}$. If four points have these pairwise distances (with the same multiplicities), are they necessarily arranged in a square with side 1?
On
As it turns out, this problem in one dimension is known as the turnpike problem. The following is an example of two finite sets of points in $\mathbb{R}$ that have the same multisets of pairwise distances: $$ \{0, 1, 2, 6, 8, 11\} \text{ and } \{0, 1, 6, 7, 9, 11\} $$ They are not congruent. Clearly, this means that the problem in dimension 2 also has a negative answer.
Yes. Place the first point arbitrarily. The second can go anywhere on the circle around that point at the appropriate distance. If all points are collinear, the distances from the first two points uniquely characterize all the other points. If not, there is just one additional freedom: the first point not collinear to the first two has two possible positions, related by reflection across the line joining the first two points. Every other point is then uniquely determined by its distances to those three points.
Thus given two such configurations, we can make them coincide with a translation (to make the first points coincide), followed by a rotation (to make the second points coincide, and then possibly a reflection across the line joining the first two points.
EDIT: Ah, now I see: the answer then seems to be no. If my computations are correct, there are two noncongruent planar configurations of four points with edge lengths $$ 1, 1, 1, 1/50, s, t$$ where $s \approx .9826294908 $ and $t \approx .9828330051$ satisfy $$\eqalign{s&={\frac{1}{50}}\,\sqrt {2500\,{t}^{2}-1}\cr 6250000\,&{t}^{4}-12507500\,{t}^{2}+6250003=0\cr} $$
In one case the triangles incident on the $1/50$ edge have their other edge lengths $1, 1$ and $1,t$; in the other case, they have $1,s$ and $1,t$.
Here is a picture of the two cases:
EDIT: This was the case $\epsilon = 1/50$ of a more general solution: for $0 < \epsilon < 1$ we can take the edge lengths to be $1, 1, 1, \epsilon, s, t$ where
$$ \eqalign{s&=\sqrt {1+\frac12\,{\epsilon}^{2}-\frac12\,\sqrt {12\,{\epsilon}^{2}-3\,{\epsilon}^{4 }}}\cr t&=\sqrt {1+\frac{3}{2}\,{\epsilon}^{2}-\frac12\,\sqrt {12\,{\epsilon}^{2}-3\,{\epsilon}^{4 }}}\cr}$$