Consider the ring $A_{p,n} = \mathbb{F}_p [x]/ (x^{p^n}-x)$. It has a basis $\{1, x, x^2, \ldots, x^{p^n - 1}\}$. The Frobenius endomorphism $x \mapsto x^p$ permutes elements of this basis. I've made the following observation:
The number of cycles of this permutation is equal to the number of distinct irreducible factors of $x^{p^n} - x$ (which itself is the product of all irreducible polynomials of degree $d$ for all $d|n$). In other words, the dimension of the fixed frobenius sub-algebra is equal to the number of distinct irreducible factors of $x^{p^n} - x$.
Moreover, the number of cycles of length $d$ is equal to the number of irreducible factors of $x^{p^n} -x$ of degree $d$.
For instance, for $A_{3,2}$ we have the frobenius endomorphism as the permutation $(1)(x \; x^3)(x^2 \; x^6)(x^4)(x^5 \; x^7)(x^8)$, $x^9-x$ factors into $x(x+1)(x+2)(x^2+1)(x^2 +x + 2)(x^2+2x+2)$.
But I only have examples of this happening - how can I prove it or find a counter example? That is, what is the connection between the cycles of the ``frobenius permutation'' and the factors of $(x^{p^n}-x)$ over $\mathbb{F}_p$ ?