Does there exist a natural number $N$ that is a power of $2$ whose digits (in the decimal representation) can be permuted to a different power of $2$?
Thoughts:
If such a number $N$ exists, then after permutation a new number $N'$ is obtained. Let us assume, without loss of generality, that $N\lt N'$.
Now since, N' and N have the same number of digits, then $N'$ must be either equal to $2N$, $4N$, or $8N$, because on multiplying $16$ clearly the number of digits increases.
Therefore $N'-N$ must be of the form $N,3N,7N$
This doesn't seem to lead anywhere though.
Hint: So we have a number $2^a$. When we permute the digits, we get say $2^b$. Note however that if we take a number $N$ and permute its digits to get $N'$, then $9$ must divide $N-N'$.
Remark: This only deals with the problem if we do not allow initial $0$'s.