Can you determine set of points from their point-to-point vectors? Let P be set of points in $R^3$. 
Given all possible point-to-point vectors $V=\{u-v : u,v\in P \}$ can you determine P up-to translation and rotation? 
Anyone know thereoms for similar problems? 
EDIT: In fact V supposed to be a sequence of all difference vectors in some order not a set. (When V is a set it is not possible see celtschk's answer below)
 A: It's not possible in general. For example, consider the two point sets
$$P_1 = \{(1,0,0), (-1,0,0), (\frac12,\frac12\sqrt{3},0), (-\frac12,\frac12\sqrt{3},0), (\frac12,-\frac12\sqrt{3},0), (-\frac12,-\frac12\sqrt{3},0)\}$$
and
$$P_2 = P_1\cup\{(0,0,0)\}.$$
That is, the first one is six points in a regular hexagon around the origin in the $xy$-plane, and the second set is that same hexagon with the origin added.
Then it is not hard to check that for both sets, you get the same set of vectors, as any vector from the origin to one of the corner points is already the vector corresponding to a side.
Since the two sets have a different number of points, they clearly don't differ only by translation or rotation.
A: Since @celtschk already gave you a nice answer about why your problem is, in the general case, impossible, I'll comment about why you can always (theoretically) find good candidates collections $P$.
Since you don't care about translations, you can always assume that the origin is part of $P$. Say your set $V$ contains $N$ vectors, just consider the $N$ points defined as translations of the origin by one of these vectors. You end up with $N$ points $\mathscr P=\{M_1,\ldots,M_N\}$.
Any good collection of points $P$ is necessarily a subset of $\mathscr P$. From there a simple (and inefficient) method consists in going through every subsets of $\mathscr P$, computing its point-to-point set of vectors, and verifying if it coincides with $V$.
Assuming that there indeed exists a collection of points $P$ that yields the set of vectors $V$, you can find every collection of points that yield that set $V$. In other words, if you define 
$$\phi (P)=\{u-v : u,v\in P\}$$
you can always compute $\phi^{-1}(V)$ (up to translation).
