Multiples and decimal expansion... If $d \in \{1;\cdots;9\}$ and $x \in \mathbb{N^*}$, $N_d(x)$ denotes the number of digits equal to $d$ in the decimal expansion of $x$.
If $n$ and $m$ are in $\mathbb{N^*}$, is there $a$ in $\mathbb{N}^*$ such that : $\forall d \in \{1;\cdots;9\}$ , $N_d(an)=N_d(am)$ ?
 A: Not completely rigorous, but let $m$ be the concatenation of to copies of $n$.  Then (ignoring carries at the joint) there will be twice as many of each digit in $am$ as there are in $an$
A: It is widely believed that there are arbitrarily large primes $p$ such that the decimal expansion of $1/p$ has period $p-1$. 
Given $m$ and $n$ with $m\gt n$, choose such a prime $p\gt10^d$, where $m$ has $d$ more digits than $n$. Now let $a=(10^{p-1}-1)/p$. Then $mp$ will have the same number of each nonzero digit as $np$, because it will have the same digits as $np$, only shifted cyclically, with some additional zeroes.  
Examples: 


*

*$m=2$, $n=1$. We can take $a=(10^6-1)/7=142857$. $am=285714$. 

*As in my comment on another answer, take $m=11$, $n=1$. We can take $a=(10^{16}-1)/17=588235294117647$. Note $d=1$. Then $am=6470588235294117$. 
I may not have everything exactly right, but I expect that using $a=(10^{p-1}-1)/p$ for some appropriate $p$ will always work (conditional on the conjecture that we never run out of full-period primes). 
A: Generally no. For example $N_0(a \times 1) < N_0(a \times 1000)$ for  $a \in \mathbb N^* $.
But this is not counterexample.
