How can I find all the fixed points of the endomporphism $f \mapsto f^p$ in the ring $ \mathbb{F}_p/ (x^{p^n} - x)$.
My hunch is that it should have something to do with $x^{p^n}-x$ being the product of all irreducible polynomials of degree d, where $d$ various over divisors of $n$, but I've not been able to find a general relationship.
(as this is not a domain, the fixed points will be more than just $\mathbb{F}_p$. For instance, for $p=n=2$, the fixed points are spanned over $\mathbb{F}_2$ by $x^2 +x$, $x^3+x^2+x$, and $x^3+1$)
The Frobenius endomorphism fixes the prime subring (i.e. the scalars) pointwise. Furthermore,
$$\left(\sum_{i=0}^{p^n-1}c_ix^i\right)^p=\sum_{i=0}^{p^n-1}c_ix^{ip} $$
(the Freshman's binomial theorem in characteristic $p$). Powering by $p$ is a permutation (in fact an automorphism) of the cyclic multiplicative group $\langle x\rangle=\{x,\cdots,x^{p^n-1}\}$ of order $|\langle x\rangle|=p^n-1$ and having identity element $x^{p^n-1}$ (the inverse is $a\mapsto a^{p^{n-1}}$). Therefore the fixed points are determined by coefficients which are constant on trajectories (cycles, in permutation lingo) of this powering-by-$p$ permutation on $\langle x\rangle$, ignoring the constant coefficient. These polynomials will be linearly spanned by those which correspond to characteristic functions of the cycles of this permutation.
Therefore, an $\Bbb F_p$-basis for the frobenius' fixed points is
$$\{1\}\cup\left \{\sum_{k\in T(p,r)} x^{k}:0<r\le p^n-1\right\}. \tag{$\circ$}$$
where $T(p,r)=\{r,pr,p^2r,\cdots\}$. The reason we do not notate the summation in the usual way, i.e. with summands $x^{p^mr}$ for $m=0,\cdots,n-1$ and $0<r\le p-1$, is that for certain choices of $r$ there can be repetitions that occur and annihilate important terms in the characteristic $p$ environment.
In particular for $p^n=2^2$ the correct set is $\{1,x+x^2,x^3\}$ as per $(\circ)$ but with the naive summation notation it would be $\{1,x+x^2+x,x^2+x+x^2,x^3+x^3+x^3\}=\{1,x,x^2,x^3\}$, incorrect.