When does linear polynomials commute over finte field? I have posted a question yesterday in When does degree 1 polynomial commute?. Which describes about  any linear polynomial $ax+b$ commute(in the sense of compostion) with  other  polynomial of degree $>1$ except identity polynomial($a=1,b=0$).
ie: Can it $f(ax+b)=af(x)+b$  happen!
I thanks to author "emacs drives me nuts" for the valuable comment.Your proof was  helpful for $f(x)\in\mathbb{R}[x]$. But I am curious to know over finite field of order $q=p^k,k>0$.
Let $f(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0\in\mathbb{F}[x]$.
If $f(ax+b)=af(x)+b$, then leading coefficient of $x^n$ is $c_na^n=ac_n$
implies
\begin{equation}
    a^{n-1}=1 
\end{equation}
since $c_n\neq 0$.
Next looking coefficient of $x^{n-1}$ which is $nc_na^{n-1}b+c_{n-1}a^{n-1}=ac_{n-1}$
substituting $a^{n-1}=1$ we get
\begin{equation}
    nc_{n}b+c_{n-1}=ac_{n-1}
\end{equation}
Comparing the constant term
which is
\begin{equation}
    f(b)=b+ac_0
\end{equation}
After that I couldn't proceed. please give me a hint to complete the proof!
 A: Listing my  findings. We want to classify the polynomials $f(x)\in\Bbb{F}[x]$ such that $f\circ h=h\circ f$, where $h(x)=ax+b\in\Bbb{F}[x]$ has degree $1$ (so $a\neq0$). I can do the cases $b=0$ and $a\neq1$, but the case $h(x)=x+b, b\neq0$, is still in the dark. Those cases I can reduce to the case $h(x)=x+1$, so settling that one would suffice! Hopefully we can resolve that later :-)
Remember that every non-zero element of a finite field is a root of unity. We first deal with the case $h(0)=0$, when $h(x)=\omega x$, and $\omega$ is a root of unity of order $m$. We know that $\gcd(m,p)=1$.
Case 1. If $\omega=1$ then, of course, every polynomial $f(x)$ works. Listing this just for the sake of completeness.
Case 2. Another easy case is $h(x)=\omega x$, $\mathrm{ord}(\omega)=m>1$. If $f(x)=\sum_i a_ix^i$, then $f(h(x))=\sum_i a_i \omega^ix^i$ and $h(f(x))=\sum_i\omega a_i x^i$. These are equal if and only if $\omega^i=\omega$ whenever $a_i\neq0$. We can conclude that $f\circ h=h\circ f$ if and only if $a_i=0$ unless $i\equiv 1\pmod m$. This is an extension of the comment by TonyK under the earlier version of the question.
Moving on, we look at the cases with a non-zero constant term. The general trick is the following.
Fact. If $g(x)$ has an inverse (with respect to the composition of polynomials), then the polynomial $f$ commutes with $h$ if and only if $g\circ f\circ g^{-1}$ commutes with $g\circ h\circ g^{-1}$.
Proof. This follows from the associativity of the composition of polynomials. QED
Consider the case of $h(x)=\omega x$, $\omega\notin\{0,1\}$ and $g(x)=x+c$. We have $g^{-1}(x)=x-c$, and
$$(g\circ h\circ g^{-1})(x)=\omega(x-c)+c=\omega x+(\omega-1)c.$$
So the choice $c=b/(\omega-1)$ yields $(g\circ h\circ g^{-1})=\omega x+ b$. Therefore, by Case 2, we have
Case 3. The polynomials $f(x)$ that commute with $h(x)=\omega x+b$, $\omega$ of order $m>1$, are the polynomials of the form
$$f(x)=c+\sum_{i\equiv1\pmod m}a_i(x-c)^i,$$
where $c=b/(\omega-1)$.
It is worth observing that this is a generalization of the classification user Emacs drives me nuts gave
in the earlier version dealing with $\Bbb{R}$ in place of $\Bbb{F}$. There $m=2$, $\omega=-1$ and $c=-b/2$.
We can milk a bit more out of the Fact. Let $g(x)=bx$, $b\neq0$, when $g^{-1}(x)=b^{-1}x$. With $h(x)=x+1$ we get
$$
(g\circ h\circ g^{-1})(x)=x+b.
$$
In view of the Fact the remaining case of $h(x)=x+b$ is thus reduced to handling the case $h(x)=x+1$. Unlike in the case of reals we have solutions $f$ of degree $>1$. The obvious one is the Frobenius automorphism $f(x)=x^p$ and its (compositional) iterates $f(x)=x^{p^m}$. This should be doable, but another idea is needed.
A: This answer fills in the missing case in Jyrki Lahtonen's answer, i.e. for $b \in \mathbb{F}_q$ we classify the $f(x) \in \mathbb{F}_q[x]$ such that $f(x+1) = f(x)+1$.
By considering $g(x) = f(x) - x$, it suffices to find all $g(x)$ such that $g(x+1) = g(x)$.
Then $g(0) = g(1) = g(2) = \cdots = g(p-1)$, so the polynomial $g(x)-g(0)$ is divisible by $\prod_{i=0}^{p-1}(x-i) = x^p -x$, i.e.
$$g(x) = g(0) + (x^p-x)h(x)$$
for some $h(x) \in \mathbb{F}_q[x]$. Since $g(x+1) = g(x)$, we must also have  $h(x+1) = h(x)$. Recursively applying the same argument, we see that $g(x)$ can be written as a polynomial in $x^p-x$, i.e. $g(x) \in \mathbb{F}_q[x^p-x]$. Conversely, any element of $\mathbb{F}_q[x^p-x]$ is invariant under the translation $x \mapsto x+1$.
So $f(x+1) = f(x)+1$ if and only if
$$f(x)-x = c_m(x^p-x)^m + \ldots + c_1(x^p-x) + c_0$$
for some constants $c_i \in \mathbb{F}_q$.
The case $f(x+b) = f(x)+b$ reduces to the above case by considering that $\hat{f}(x) := b^{-1}f(bx)$ satisfies $\hat{f}(x+1) = \hat{f}(x)+1$.
