Overlaying a point cloud on a two-dimensional probability distribution such that the local point density corresponds to the local probability density Say I sample a set of points $(p_1,p_2,...) \in P$ from a probability distribution $f(x,y)$, e.g. a bivariate normal distribution, such that my sampling process chooses a point in the distribution with a probability proportional to the value $f(x,y)$ at that point.  In particular, let's say we use the Metropolitan-Hastings algorithm to sample the set of points $P$ via a biased random walk.

Now, I take $P$, and I apply some arbitrary geometric rotational and translational geometric transform that acts uniformly on all points $p_i \in P$.  I provide this transformed set of points $P^*$ to you, and your job is to reverse the geometric transform and map the coordinates back to their original positions to the best of your ability.

What is the optimal way to proceed?

A thought is that we can sample a new set of points $P_2$ from the distribution $f(x,y)$, and then use something like the RANSAC algorithm (http://en.wikipedia.org/wiki/RANSAC) to find a geometric transform to align the $P^*$ and $P_2$.  This, however, seems to be a bit wasteful?

Motivation - Say we know that the maxima of the distribution $f(x,y)$ occurs at some $(x_k,y_k)$.  What is $(x_k,y_k)^*$, i.e. where is this point after applying the geometric transform that mapped $P$ to $P^*$?  We could also ask about other "distinct" points of interest in $f(x,y)$ so that it isn't simply a matter of computing something like a geometric mean (or perhaps geometric median) of the point set $P^*$ to calculate $(x_k,y_k)^*$.  Can we figure this out without the kind of silly solution I propose the employs the RANSAC algorithm?

(In response to R. B.'s comments): I mean "up to symmetry" here - apologies for the confusion.
Overlaying a point cloud on a two-dimensional probability distribution such that the local point density corresponds to the local probability density is a good way to define the cell generators of a centroidal Voronoi tessellation (CVT).

CVTs don't loose their quantization precision by translation or rotation in concert with the underlying probability distribution. Not sure what your question is.