Cut a $4 \times 4 \times 4$ cube into 64 $1 \times 1 \times 1$ small cubes, then dye 16 of the $1 \times 1 \times 1$ small cubes red, requiring that among any 4 small cubes parallel to any edge, exactly 1 small cube is dyed red. The number of different dyeing methods is $\qquad$ (dyeing methods that are the same after rotation are also considered different dyeing methods).
A $4 \times 4 \times 4$ cube, from top to bottom, has 4 layers, each a $4 \times 4 \times 1$ rectangular prism; according to the problem, "in any 4 small cubes parallel to an edge, exactly 1 small cube is painted red," so in each $4 \times 4 \times 1$ layer, exactly 4 small cubes are painted red;
and when viewed from above, the red small cubes in each layer form a $4 \times 4$ square.
For this $4 \times 4$ square, we use “1”, “2”, “3”, “4” each 4 times,
to represent the small squares in the 1st, 2nd, 3rd, and 4th layers that are painted red;
and according to the problem, in this $4 \times 4$ square, each row and each column contains “1”, “2”, “3”, “4” exactly once.
Thus, the problem is transformed into "how many different ways are there to fill a $4 \times 4$ square with “1”, “2”, “3”, “4” each 4 times, such that each row and each column contains “1”, “2”, “3”, “4” exactly once?"
First, consider “1”, with one in each row and in different columns, there are $4 \times 3 \times 2 \times 1=24$ ways to fill; let's assume it is in the configuration of Figure 1.
Next, consider “2”, the “2” in the first row has 3 ways to fill; let's assume it is in the configuration of Figure 2;
If the “2” in the second row is in the first column, then the “2” in the third row is in the fourth column, and the “2” in the fourth row is in the third column; (as in Figure 3)
If the “2” in the second row is in the third column, then the “2” in the fourth row is in the first column, and the “2” in the third row is in the fourth column; (as in Figure 4)
If the “2” in the second row is in the fourth column, then the “2” in the third row is in the first column, and the “2” in the fourth row is in the third column; (as in Figure 5)
Next, consider “3”, the “3” in the first row has 2 ways to fill; let's assume the “3” in the first row is in the third column (as in Figures 6 and 8).

Figure 6 has 2 remaining ways to fill; (as in Figures 9 and 10)
Figure 7 has 1 remaining way to fill; (as in Figure 11)
Figure 8 has 1 remaining way to fill; (as in Figure 12)
Thus, if the “2” in the first row is in the 2nd column, and the “3” in the first row is in the 3rd column, there are 4 ways;
The total number of different ways is $24 \times 3 \times 2 \times 4=576$.
The original problem has 576 different coloring methods.