a.) Let us see what happens if we have a two digit number. Then there are 102 = 100 two digit integers and only 10 of them are non-equivalent, namely (00, 11, 22, 33, 44, 55, 66, 77, 88, 99). The other 90 two digit integers are equivalent. For example 01 is equivalent to 10 and 02 to 20 and so on.
Now take the 103 = 1000 three digit integers. Only 10 of these are non-equivalent as well, namely (000, 111, 222, 333, 444, 555, 666, 777, 888, 999). The other 990 are equivalent.
Extending this to the six digit integers we see that of the 106 six digit integers 10 of them are non-equivalent and the other 106 - 10 are equivalent.
b.) If 0 and 9 can appear at most once in a six-digit integer then there are only 8 non-equivalent integers, namely (111111, 222222, 333333, 444444, 555555, 666666, 777777, 888888).
c.) Generalizing a.) we see that if we have an n digit integer then of the 10n n digit integers once again only 9 of them are non-equivalent and the other 10n - 9 are equivalent.
Generalizing b.) we see that there are only 8 non equivalent integers among the 10n n-digit integers, namely (111...., 222...., 333...., 444...., 555...., 666...., 777....., 888....).
Hope this helps!
Jonathan