Around a circle, 12 fields are arranged. Among these, four adjacent fields are occupied by 4 different figures: a red, a yellow, a green, and a blue one in sequence. Each figure can be moved by jumping over four adjacent fields to land on the fifth, provided that this field is not already occupied by another figure.

After a certain number of moves, the figures return to the four fields they originally occupied, but in a different order. In how many different orders can they return to their original positions?
For now, let's not worry about the rule that a figure cannot move over a square occupied by another figure, in which case 12 (one-way) steps will touch every square once. A similar but simpler case can be achieved if we modify the task so that instead of jumping over four squares, each figure steps to an adjacent square. The solution to the modified problem is clear: considering the prohibition of stepping on the same square, the figures can return to their original positions in any cyclic order, which can happen in four ways.
![](https://cdn.mathpix.com/cropped/2024_05_02_9dd4e75dc4dbafba5524g-1.jpg?height=370&width=734&top_left_y=354&top_left_x=684)

The original problem can then be reduced to the modified problem by changing the order of the squares according to the steps of one of the figures, for example, the red figure, as shown in the second diagram. When a figure on the first diagram steps to the fifth square by skipping four squares in a certain direction, the same figure on the second diagram steps to an adjacent square, and vice versa. Therefore, if we place the four figures on the 1st, 2nd, 3rd, and 4th squares of the second diagram, our task will be to move these figures to the same squares in a different order in a certain number of steps, always stepping to an adjacent square, provided that no figure is already standing there. It is clear that this can also happen in four ways, without changing the cyclic order of the figures.

Note. The same reasoning can be used to prove that if there are $n$ squares on the circumference of a circle and $k$ different figures stand on $k$ of these squares, and each figure can be placed on the $m$-th square by skipping $m-1$ squares (if no figure is already standing there), the figures can also return to their original squares in a cyclic order, i.e., in $k$ ways, provided that $n$ and $m$ are relatively prime. Only under this latter condition can every figure reach any square during the steps, which we utilized in the proof. If $n$ and $m$ are not relatively prime, the squares will split into multiple groups, and the steps of figures belonging to different groups do not affect each other. For example, if $m=2$ in the originally set problem, a figure standing on an even square will never reach an odd square and vice versa. In such a case, the cyclic order of figures belonging to each group remains unchanged separately.