Profile Matching in a Point Cloud
This is always going to require a lot of computation, especially if you want to process as many as 2000 points.  I'm sure there are highly-optimized solutions for this kind of pattern-matching already, but you have to figure out what it's called in order to find them.  

Since you're talking about a point cloud (sparse data) instead of an image, my cross-correlation method doesn't really apply (and would be even worse computationally).  Something like RANSAC probably finds a match quickly, but I don't know much about it.

My attempt at a solution:

Assumptions:


You want to find the best match, not just a loose or "probably correct" match
Match will have some small amount of error due to noise in measurement or computation
Source points are coplanar
All source points must exist in target (= any unmatched point is a mismatch for the entire profile)


So you should be able to take a lot of shortcuts by disqualifying things and decrease computation time.  In short:


pick three points from source
search through target points, finding sets of 3 points with the same shape
when a match of 3 points is found, check all the other points in the plane they define to see if they're a close match
if more than one match of all points is found, choose the one with the smallest sum of 3D distances error


More detailed:

pick a point from the source for testing s1 = (x1, y1)
Find nearest point in source s2 = (x2, y2)
d12 = (x1-x2)^2 + (y1-y2)^2
Find second nearest point in source s3 = (x3, y3)
d13 = (x1-x3)^2 + (y1-y3)^2
d23 = (x2-x3)^2 + (y2-y3)^2

for all (x,y,z) test points t1 in target:
    # imagine s1 and t1 are coincident
    for all other points t2 in target:
        if distance from test point &gt; d12:    
            break out of loop and try another t2 point
        if distance ≈ d12:
            # imagine source is now rotated so that s1 and s2 are collinear with t1 and t2
            for all other points t3 in target:
                if distance from t1 &gt; d13 or from t2 &gt; d23:
                    break and try another t3
                if distance from t1 ≈ d13 and from t2 ≈ d23:
                    # Now you've found matching triangles in source and target
                    # align source so that s1, s2, s3 are coplanar with t1, t2, t3
                    project all source points onto this target plane 
                    for all other points in source:
                        find nearest point in target
                        measure distance from source point to target point
                        if it's not within a threshold:
                            break and try a new t3
                        else:
                            sum errors of all matched points for this configuration (defined by t1, t2, t3)


Whichever configuration has the least squared error for all other points is the best match

Since we're working with 3 nearest neighbor test points, matching target points can be simplified by checking if they are within some radius.  If searching for a radius of 1 from (0, 0), for instance, we can disqualify (2, 0) based on x1 - x2, without calculating the actual Euclidean distance, to speed it up a bit.  This assumes that subtraction is faster than multiplication. There are optimized searches based on more arbitrary fixed radius, too.

function is_closer_than(x1, y1, z1, x2, y2, z2, distance):
    if abs(x1 - x2) or abs(y1 - y2) or abs(z1 - z2) &gt; distance:
        return False
    return (x1 - x2)^2 + (y1 - y2)^2 + (z1 - z2)^2 &gt; distance^2 # sqrt is slow


3D Euclidean distance is $d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2+(z_1 - z_2)^2}$
but square roots are slow. Since we're just comparing relative distance, I think we could drop the sqrt and consider "distance" to be sums of squares.

The minimum calculation time would be if no 2-point matches are found.  If there are 2000 points in the target, this would be 2000 * 2000 distance calculations, though many would be disqualified by a subtraction, and the results of previous calculations could be stored so you only have to do $2000 \choose 2$ = 1,999,000.

Actually, since you will need to calculate all of these anyway, whether you find matches or not, and since you only care about nearest neighbors for this step, if you have the memory it's probably better to pre-compute these values using an optimized algorithm.  Something like a Delaunay or Pitteway triangulation, where every point in the target is connected to its nearest neighbors.  Store those in a table, then look them up for each point when trying to fit the source triangle to one of the target triangles.

There's a lot of calculations involved, but it should be relatively fast since it's only operating on the data, which is sparse, instead of multiplying lots of meaningless zeros together like cross-correlation of volumetric data would involve.  This same idea would work for the 2D case if you found the centers of the dots first and stored them as a set of coordinates.