Source: Reader Sudipta Das.
Joseph DeVincentis' solution is representative of most solutions:
Part 1: Can there exist a (multi-digit) number, consisting of only distinct digits, all of whose "rotations" (including the number itself) are exact multiples of each of its digits? Such a number cannot contain 0, since all multiples of 0 are 0. Such a number cannot contain 5, since all multiples of 5 end in 5 or 0, and we already know the number cannot contain 0. If such a number contains any even digit, then all the digits in the number must be even. So, we are left with combinations of 2, 4, 6, 8 and combinations of 1, 3, 7, 9. If the number contains 4 or 8, then it cannot contain 2 or 6, because at some point in the number a 2 or 6 must follow a 4 or 8, and numbers that end in 42, 46, 82, or 86 are not multiples of 4. And likewise 48 and 84 do not work because 84 is not a multiple of 8. The numbers 26 and 62 do not work because they are not multiples of 6. If the number contains 3 or 9, it cannot contain 1 or 7 because multiples of 3 always have digits which add to a multiple of 3, but the combinations of 1, 3, 7, and 9 which contain a 1 or 7 have digit sums which = 1 or 2 mod 3. The numbers 39 and 93 do not work because they are not multiples of 9. Thus, no such number exists. Part 2: Can there exist a number, consisting of at least 2 distinct digits (repetition of digits allowed), all of whose "rotations" (including the number itself) are exact multiples of its distinct digits? Some of the reasoning from above still works, but there are now a few cases where we can create a number with this property: 2226, and other combinations of 2s and 6s with a multiple of three 2s. 1113, and other combinations of 1s and 3s with a multiple of three 1s. 3339, 111339, 1111111119, and other combinations of 1s, 3s, and 9s with a digit sum that is a multiple of 9, including cases with no 1s and no 3s. Some combinations of 7s with other odd digits also work. For example, keeping in mind that 1001 = 7 * 11 * 13, numbers like 117117 and 177177 work.
Now the odd case:
1113 (or any number with 3k ones and at least one three.)
3339 (or any number with 3k threes and at least one nine.)
1111111119 (or any number with 9k ones ant at least one nine.)
111339 (basically if using (1,3,9) and the first digits sum to 9, the rotations will all work out.)
373737 (any 6k digit number divisible by 7 will rotate to a number divisible by 7.)
717913 (uses all possible odds.)
As you've pointed out, 5 cannot be in the number.
7 can only be used in a 6k digit number, so repetition of digits is required if a 7 is involved.
This leaves only 1, 3, and 9. We can't use them in a 3-digit number (the sum of the digits shows we won't get a number divisible by 9), and we can see the two-digit numbers won't work. So repetition is required and some examples are above.