Nazar_Serdyuk wrote: Let $n, a, b$ be integers with $n \geq 2$ and $a \notin \{0, 1\}$ and let $u(x) = ax + b$ be the function defined on integers. Show that there are infinitely many functions $f : \mathbb{Z} \rightarrow \mathbb{Z}$ such that $f_n(x) = \underbrace{f(f(\cdots f}_{n}(x) \cdots )) = u(x)$ for all $x$. If $a = 1$, show that there is a $b$ for which there is no $f$ with $f_n(x) \equiv u(x)$. We apply $f$ to both sides to get $\begin{*align}f_{n+1}(x)=f_{n+1}(x)\implies f_n(f(x))=f(f_n(x))\implies u(f(x))=f(u(x))\implies f(ax+b)=af(x)+b\;\;\;(\star)$. Now we claim that there are infinitely many linear functions that work. Let $f(x)=rx+s$ It suffices to show that $\exists$ infinitely many $r,s\in\mathbb{Z}$ satisfying $(\star)$ .Now comparing both sides we need $br+s=as+b\implies br-s(a-1)=b$. Now this diophantine equation has a solution if $\text{gcd}(b,a-1)\mid b$.But that's trivially true.So this diophantine equation has infinitely solutions in $\mathbb{Z}$.And we are done.

That's not at all true. Where do you use $n$ in your answer? The usual solution is to (except for $a=0, \pm 1$) is to divide those elements not in the range of $u$ (of which there are an infinite number) into $n$ equinumerous sets, with bijections from the first to the second to the third ... to the last to the first again. There are certainly an infinite number of ways to do this. Then our $f$ is equal to these bijections except for the last one: in this case, it's equal to $u$ composed with the bijection. And $u(f(x))=f(u(x))$, extending $f$ to the range of $u$. Since every element in $\mathbb{Z}$ can be taken back a finite number of times, every element of $\mathbb{Z}$ has $f$ defined on it. Now for $a=-1$, we can divide the set of pairs $\{x, b-x\}$ into $n$ equinumerous sets with bijections from each set to the next, and from the last to the first. And again, take $f$ to be the bijections from one element in the first pair to an element in the second pair, from that to an element in the third pair, and so on: then $f$ is $u$ composed with the final bijection, and $f(u(x))=u(f(x))$. If $b$ is even, we require $f\left(\frac{b}{2}\right)=\frac{b}{2}$. If $a=0$ then it doesn't work: $f(x)=b$ is clearly the only function that works. If $a=b=1$ then it also doesn't work: as joybangla wrote, $f(x+1)=f(x)+1$ so $f(x)=x+c$ for some $c$. But this gives us $u(x)=f_n(x)=x+nc\neq x+1$.