Can a collection of points be recovered from its multiset of distances? Consider $n$ distinct points $x_1,\dots,x_n$ on $\mathbb{R}$. Associated to these points is the multiset of all distances $d(x_i,x_j)$ between two points. Suppose one is only handed this multiset (you do not know the corresponding indices). Does this allow one to uniquely recover the original points up to reflection and translation? 
 A: 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
A: Here's a conceptual explanation for non-uniqueness for $n \geq 7$.
Steve's first link, Lemka--Skiena--Smith's "Reconstructing Sets From Interpoint Distances", relies on Rosenblatt--Seymour's earlier "The Structure of Homometric Sets". Rosenblatt--Seymour's main innovation is the following observation. First, if U and V are finite multisets in $\mathbb{R}^n$, then $U+V$ and $U-V$ have the same multiset of pairwise differences. Second, the "virtual converse" holds: if $A$ and $B$ have the same multiset of pairwise differences, then there are "virtual multisets" $U$ and $V$, i.e. multisets where we allow negative multiplicities, such that $A = U+V$ and $B = U-V$.
In terms of generating functions, let $A(x) := \sum_{a \in A} x^a$. $A$ and $B$ having the same multiset of pairwise differences means precisely that $A(x)A(x^{-1}) = B(x)B(x^{-1})$, and the Rosenblatt--Seymour criterion means exactly that there are $U(x), V(x)$ with integer coefficients such that $A(x) = U(x)V(x)$ and $B(x) = U(x)V(x^{-1})$. (Here the exponents are real numbers, and it's multivariate in the case of $\mathbb{R}^n$.) In any case, the insight is that we can factor $A(x)$ into irreducibles and toggle $x \mapsto x^{-1}$ in a subset of the factors to construct all possible multisets with the same distances. Here factorization is unique up to units $\pm x^\alpha$.
Uniqueness with at most 5 points up to translation and reflection is hence equivalent to the following claim. If $U(x)V(x)$, $U(x)V(x^{-1})$, and $U(x^{-1})V(x)$ all differ by more than just overall x-shifts and all have non-negative integer coefficients, then the common sum of their coefficients is at least 6. There's of course some minimum value for this common sum, though it's not at all obvious that it's exactly 6. You can let U and V be non-symmetric 0-1 polynomials to immediately get common sums of the form $nm$ for $n, m \geq 3$. In fact, you can use $U(x) = 1-x+x^3$ and let $V(x)$ be any non-symmetric 0-1 polynomial with integer exponents and non-zero coefficients occurring in clumps of at least 3 adjacent to get all numbers $\geq 7$. Getting $n=6$ in this way seems difficult. Some brute force gives for instance
$$x^{32}+x^{22}+x^{20}+x^9+x^1+1 = \left(x^2+x+1\right) \left(x^9-x^8+x^6-x^5+x^3-x^2+1\right) \left(x^{21}-x^{10}+x^9+1\right)$$
Perhaps azimut's arguments can be recast into the generating function setup and made more intuitive?
