Let's assume there is two powers of 2 $2^m$ and $2^n(m<n)$ such that one can permute decimal digits of the 1st to obtain 2nd. Then $n-m\leq 3$, else $2^n\geq 2^{m+4}>10*2^m$ and $2^n$ has more decimal digits, than $2^m$. Hence $2^n-2^m=2^m(2^{n-m}-1)=2^m$, $3*2^m$ or $7*2^m$ is not divisible by 9, and $2^m\not\equiv2^n(mod9)$. And since $S(2^m)\equiv2^m(mod9)$, $S(2^n)\equiv2^n(mod9)$ we have $S(2^m)\not\equiv{S(2^n)}(mod9)$ $S(2^m)\not=S(2^n)$ we can't permute decimal digits of $2^m$ to obtain $2^n$.

P.S. $S(x)$ is the sum of decimal digits of $x$.

What if N has zeros in its decimal form and those zeros when permuted are placed in front so the result of permutation can decrease more than 10 times?