OK, so I'm really loosing faith on the 4x3-counting.

I just had the ideas on how to attac it, but I didn't do enough on the estimates. I now see that the task is too big.

There are 12!/(4!*4!*4!*3!)=5775 ways to group 12 symbols into an unordered tripple of disjoint 4-element sets. That means that the number of band configurations where the columns within a box is ordered by their smallest element, and where the configuration of the first box is C0={{1,2,3,4},{5,6,7,8},{9,10,11,12}}, is 5775^3.

Given a band configuration on this form, there are 24^4*6 operations that will lead to all the other elements in the same equivalence class: 4! ways to permute the boxes and 3! ways to permute the columns in the first box before the symbols are repermuted to put the first box on the base form C0, then another 4!^3 ways to permute the symbols within each column in the first box.

So each equivalence class ha a maximum size of 24^4*6, hence there are at least 5775^3/(24^4*6) = 96752 gangsters in 4x3-sudoku. As there are 346 ways to choose the column configuration of each box on the second row, the innermost loop when doing the counting must be run at least 96752 * 346^4 / 6 = 2.3e14 times. And that is not the only problem, it can be hard for a band configuration to find the way back to its gangster, or in another way find out its total number of sudoku bands - without wasting to much time.

Unless someone comes up with a clever shortcut here, I regard the exact number of 4x3-sudokus as inaccessible for us mortals.

Frustrating when the estimate I have given earlier probably is pretty close...