This solution takes about 6 seconds to run on my system.

Essentially it works as follows: Start by finding a scanner k that matches with scanner 0, and transform scanner k coordinates to scanner 0 coordinates. Next, find a scanner that either matches with scanner 0 or with scanner k, transform to scanner 0 coordinates — repeat until all scanners are matched.

For checking if two scanners match, go through all 24 orientations for either of the scanners. For each orientation, check how many overlapping beacons you get by transposing any beacon of the second scanner onto any beacon of the first scanner. Stop if there are at least 12 overlaps.

The code uses a major optimization which reduces the runtime from 20+ minutes to a few seconds: For a given pair of scanner, assume they overlap in at least 12 beacons. The distance between beacons is invariant under transposition and rotation. So the distances between these 12 beacons will be the same for both scanners. There are 66 pairwise distances between 12 beacons. So the intersection of the set of all pairwise distances for the first scanner, and the respective set for the second scanner, must contain at least 66 elements.

So we can skip a deep comparison for any pair of scanners with less than 65 elements in the pairwise distance intersection. For my specific input, this heuristic eliminates all fruitless in-depth matching attempts.