1.5 When a piece of paper with digits is flipped upside down, the digits $0,1,8$ remain unchanged, the digits 6 and 9 swap, and other digits have no meaning when flipped. Find the number of nine-digit numbers that remain unchanged when the paper is flipped upside down.
[Solution] When the paper with a nine-digit number is turned upside down, the first digit becomes the last, the second digit becomes the eighth, and so on, with the fifth digit remaining unchanged. Therefore, the nine digits can be divided into 5 groups: the first and last digit as one group, the second and eighth digits as one group, the third and seventh digits as one group, the fourth and sixth digits as one group, and the fifth digit as its own group. It is evident that the first group has 4 different values: $(1,1),(8,8),(6,9),(9,6)$; the second, third, and fourth groups each have 5 different values: $(0,0),(1,1),(8,8),(6,9),(9,6)$; and the fifth group has three different values: 0, 1, 8. Therefore, the number of nine-digit numbers that meet the requirements of the problem is $4 \times 5^{3} \times 3=$ 1500.