Example 3 (Problem from the 22nd Moscow Mathematical Olympiad) When a paper with digits written on it is turned upside down, the digits $0,1,8$ remain unchanged, the digits 6 and 9 interchange, and other digits have no meaning when turned upside down. Find the number of 9-digit numbers that remain unchanged when the paper is turned upside down.
Solve: Divide the 9-digit array into 5 groups, the first and last digits form one group, the 2nd and 8th digits form one group, the 3rd and 7th digits form one group, the 4th and 6th digits form one group, and the 5th digit forms a group by itself. The 1st group can take 4 different values: (1,1), (8,8), (6,9), (9,6), the 2nd, 3rd, and 4th groups can take 5 different values: (0,0), (1,1), (8,8), (6,9), (9,6), and the 5th group can take three different values: 0, 1, 8. By the multiplication principle, the number of 9-digit numbers that meet the conditions is \(4 \times 5^{3} \times 3 = 1500\).