On wildness/existence of an arithmetic function. I am interested in the following problem:

Let $f:\mathbb N\to \mathbb N$ be a function and let $k$ be some fixed constant natural number more than $1$. If$$f(kx)=kf(x)$$holds for all $x\in\mathbb{N}$, what can we say about the function $f(x)$? Do there exist any non trivial solution (i.e., some solution other than identical mapping) $?$

For $k=3,$ I found kinda similar problem about existence of such a function here. 

This problem also motivates us to think about existence of function $f:\mathbb N\to\mathbb N$ for which
  $$f\circ f\circ\underbrace{\cdots}_{k~\text{compositions}}\circ f(n) = mn$$ for some fixed pair $(m,k)\in\mathbb N^2$. Does such a function exists for all values of $m$ and $k$?

The above problem is kinda similar problem as the first one as we can take one more compositions both sides in the above equation to get $mf(n)=f(mn)$. But here identical mapping won't be a solution of course! As of now, I have no idea how we can proceed for construction for such functions or do they even exists.
Any help regarding this will be highly appreciated!
 A: For the first functional equation, there exists infinitely many pathological solutions. Simply set $f(x)$ to be whatever you want for $k \nmid x$. Then, you follow the requirement and set $f(kx)=kf(x)$ for $k \nmid x$. Then, you set $f(k^2x)=kf(kx)=k^2f(x)$ for $k \nmid x$ and so on. Clearly, any positive integer can be written as $k^nx$ where $x$ is a positive integer and $k$ is a non-negative integer. You can clearly see that this construction obeys the given rule. 
For the second functional equation, simply take the set of all numbers not divisible by $m$ and divide it into $k$-tuples. Let $(x_1,x_2,\ldots, x_k)$ be such a $k$-tuple. You simply define $f(m^tx_i)=m^tx_{i+1}$ when $i<k$ and define $f(m^tx_k)=m^{t+1}x_1$. Do the same for all $k$-tuples. You can clearly see that this satisfies our requirements. You Thus, our answer is yes. 
A: Partial answer: Let $k$ be prime. Then all $n\in\mathbb{N}$ may uniquely be written as $n=k^j m$ with $j\geq 0$ and $m\in \mathbb{N}_k:=\{l\in\mathbb{N} \mid k\not\mid l\}$. Then, given any function $g\colon \mathbb{N}_k\to\mathbb{N}$, the function $f$, defined by $f(k^j m):=k^j g(m)$, satisfies $f(k n)=k f(n)$ for all $n$.
