Can a collection of points be recovered from its multiset of distances?

This is a really nice question!

Counterexample for $n=6$

The sets $$\{0,1,4,5,11,13\}\\\{0,1,2,6,10,13\}$$ are affinely inequivalent, but the multiset of differences is in both cases $$1^2\cdot 2\cdot 3\cdot 4^2 \cdot 5 \cdot 6\cdot 7 \cdot 8\cdot 9\cdot 10\cdot 11\cdot 12\cdot 13.$$

Counterexamples for $n \geq 7$ According to Lemma 2.1 in the first linked article in the answer of Steve Kass, for all $n\geq 6$, a counterexample is given by $$X\cup \{n+1,n+3\}\quad\text{and}\quad X\cup\{2,n+1\}$$ where $X = \{5,6,7,\ldots,n-1,n-2\} \cup \{0,1,n,n+5\}$.

Uniqueness for $n\leq 5$

For $n\in\{1,2\}$, the uniqueness is clear.

Let $X = \{x_1,x_2,\ldots,x_n\}$ be a point set. We assume $x_1\leq x_2\leq\ldots \leq x_n$. Up to affine equivalence, we may assume $x_1 = 0$. We denote the distance of two points $x,y$ by $\delta(x,y) = \left|x-y\right|$ and furthermore define the abbreviation $\delta_{i,j} = \delta(x_i,x_j)$. In the multiset of distances, let $a \geq b\geq c\geq d$ be the largest elements. It is clear that $a = \delta_{1,n}$ and thus $x_n = a$. Up to affine equivalence, we may assume $b = \delta_{1,n-1}$ (the other possibility is $\delta_{2,n}$), so $x_{n-1} = b$. This shows the uniqueness for $n = 3$.

For $n=4$, $\{\delta_{1,2},\delta_{2,n}\} = \{c,x_4-c\}$. This distinguishes the last remaining distance $\delta_{2,3}$, which in turn fixes $x_2$.

It remains to consider the case $n=5$.

First we see that if one further point $x\in\{x_2,x_3\}$ is fixed, the set $X$ is completely determined: Let $y$ be the missing point. Among the four remaining distances $\delta(x_1,y)$, $\delta(x,y)$, $\delta(y,x_4)$, $\delta(y,x_5)$, the maximum $d_m$ is contained in the set $\{\delta(x_1,y),\delta(x_5,y)\} = \{d_m, x_5-d_m\}$. So we also know the set $\{\delta(x,y),\delta(y,x_4)\}$. Because of $x,y\in(x_0,x_4)$, $\delta(y,x_4)$ is the larger of those two distances, which fixes the point $y$.

By the choice of $c$, we have $c = \delta_{1,3}$ (Case A) or $c = \delta_{2,n}$ (Case B). If the multiset of distances admits only one of those two cases, then by the above reasoning, $X$ is uniquely determined. So we have to see that if both cases are possible, then the point sets are necessarily identical.

Case A) If $c = \delta_{1,3}$, then $x_3 = c$, and $d\in\{\delta_{1,2},\delta_{2,5}\}$.

Case A1) $d = \delta_{1,2}$. Now $X = \{0,d,c,b,a\}$, and the $10$ distances are $$\delta_{1,2} = d,\quad \delta_{1,3} = c,\quad \delta_{1,4}= b,\quad \delta_{1,5} = a,\\ \delta_{2,3} = c-d,\quad \delta_{2,4} = b-d,\quad \delta_{2,5} = a-d,\\ \delta_{3,4} = b-c,\quad \delta_{3,5} = a-c, \\\delta_{4,5} = a-b$$

Case A2) $d = \delta_{2,5}$. Now $X = \{0,a-d,c,b,a\}$, and the $10$ distances are $$\delta_{1,2} = a-d,\quad \delta_{1,3} = c,\quad \delta_{1,4}= b,\quad \delta_{1,5} = a,\\ \delta_{2,3} = -a+c+d,\quad \delta_{2,4} = -a+b+d,\quad \delta_{2,5} = d,\\ \delta_{3,4} = b-c,\quad \delta_{3,5} = a-c, \\\delta_{4,5} = a-b$$

Case B) If $c = \delta_{2,5}$, then $x_2 = a-c$, and $d\in\{\delta_{1,3},\delta_{3,5},\delta_{2,4}\}$.

Case B1) $d = \delta_{1,3}$. Now $X = \{0,a-c,d,b,a\}$, and the $10$ distances are $$\delta_{1,2} = a-c,\quad \delta_{1,3} = d,\quad \delta_{1,4}= b,\quad \delta_{1,5} = a,\\ \delta_{2,3} = -a+c+d,\quad \delta_{2,4} = -a+b+c,\quad \delta_{2,5} = c,\\ \delta_{3,4} = b-d,\quad \delta_{3,5} = a-d, \\\delta_{4,5} = a-b.$$

By the above consideration, B1) and A1) cannot appear both (since they have $4$ points in common).

Assume that both B1) and A2) are possible. Then by comparing the distances, the two sets $\{-a+b+d, b-c\}$ and $\{-a+b+c, b-d\}$ must be the same. In both possibilities to match the elements, we end up with $c = d$, which shows that the point set is the same in both cases.

Case B2) $d = \delta_{3,5}$. Now $X = \{0,a-c,a-d,b,a\}$, and the $10$ distances are $$\delta_{1,2} = a-c,\quad \delta_{1,3} = a-d,\quad \delta_{1,4}= b,\quad \delta_{1,5} = a,\\ \delta_{2,3} = c-d,\quad \delta_{2,4} = -a+b+c,\quad \delta_{2,5} = c,\\ \delta_{3,4} = -a+b+d,\quad \delta_{3,5} = d, \\\delta_{4,5} = a-b.$$ We go on similarly as in B1).

Case B3) $d = \delta_{2,4}$. Here necessarily $a+d = b+c$, and the point $x_3$ is not yet fixed. The point set is $X = \{0,a-c,x_3,b,a\}$. We go on similarly as in B1), B2) to see that if B3) occurs together with A1) or A2), then $X$ is uniquely determined.

EDIT: The uniqueness for $n\leq 5$ is also stated in Lemma 2.1. in the first linked article of Steve Kass. However, the proof doesn't give to many details, and I do not understand the part "since $a + b = c$, if $a + c = 1$ then $b$ uniquely determines $T$.".

This problem is called the "turnpike problem" or the "partial digest problem." Sets like the two @azimut gave are called "homometric" or "homeometric," and there can be many for a given set of distances (but the number of them is always a power of two). Here are a couple of references:

Reconstructing Sets From Interpoint Distances

The Partial Digest Problem

On the Turnpike Problem