Question about the cycle type of the Frobenius Automorphism Let $f$ be an irreducible and separable polynomial of degree $n$ over $\mathbb{F}_p$. For a finite field $F$ of characteristic $p$, we define $\phi_p:\alpha\mapsto\alpha^p$. Then we know $\phi_p$ is an automorphism of $F$ and in particular we also know


*

*$\text{Gal}(f/\mathbb{F}_p)=\text{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)=\langle\phi_p\rangle$

*$\text{Gal}(f/\mathbb{F}_p)$ is isomorphic to a subgroup of $S_n$

*The order of $\phi_p$ is $n$.


But, how do we know $\phi_p$ corresponds to an element of cycle type $n$ in $S_n$ ?
Similarly, if $f$ is a separable polynomial of degree $n$ over $\mathbb{F}_p$ and
$$
f=\prod_{i=1}^r f_i
$$
where $f_i$ is an irreducible polynomial of degree $n_i$ over $\mathbb{F}_p$, we know


*

*$\text{Gal}(f/\mathbb{F}_p)=\text{Gal}(\mathbb{F}_{p^m}/\mathbb{F}_p)=\langle\phi_p\rangle$ where $m:=\text{lcm}(n_i:i=1,\ldots,r)$

*$\text{Gal}(f/\mathbb{F}_p)$ is isomorphic to a subgroup of $S_n$

*The order of $\phi_p$ is $m$.


But, how do we know $\phi_p$ corresponds to an element of cycle type $(n_1,n_2,\ldots,n_r)$ in $S_n$ ?
 A: The cycle type is determined by the way you identify $\phi_p$ as an element of $S_n$.
There is no canonical way of doing this, which makes your question a bit underdetermined, and may also be the root cause of your difficulties.
It looks like you are doing it (or, expected to do it) as follows. Let $X=\{\alpha_1,\alpha_2,\ldots,\alpha_n\}$ be the set of zeros of $f$ in its splitting field. Then $\phi_p$ will permute the elements of $X$ among themselves by the usual argument. Thus $\phi_p$ can be viewed as an element of $\operatorname{Sym}(X)$. Or, alternatively this gives us an injective homomorphism $T:\operatorname{Gal}(f/\Bbb{F}_p)\to \operatorname{Sym}(X)$. Only after this identification can we meaningfully ask about the cycle type of $\phi_p$ - namely we can ask about the cycle type of $T(\phi_p)$.
Note that separability implies that $|X|=n$, so we can think of $\operatorname{Sym}(X)$ as $S_n$.
With that out of the way we can actually answer the question. Assume first that $f$ is irreducible. Let $\beta_0$ be one of the zeros of $f$. Then, by the usual argument,
$\beta_i:=\phi_p^i(\beta_0), i=1,2,\ldots,$ are all also zeros of $f$. Because $f$ has only finitely many zeros there are repetitions among the elements $\beta_i$. If $\beta_j=\beta_i$ for some pair of indices $i<j$, then we have 
$$\phi_p^i(\beta_{j-i})=\phi_p^i(\phi_p^{j-i}(\beta_0))=\phi_p^j(\beta_0)=\beta_j=\beta_i=\phi_p^i(\beta_0).$$
As $\phi_p^i$ is bijective this implies that $\beta_0=\beta_{j-i}$. Let $\ell$ be the smallest positive integer such that $\beta_\ell=\beta_0$. By the above argument such an $\ell$ exists. Using these bits we can then easily deduce that the elements
$\beta_0$, $\beta_1$, $\ldots$, $\beta_{\ell-1}$ are all distinct. Furthermore $\phi_p$
permutes them according to the $\ell$-cycle 
$$\beta_0\mapsto\beta_1\mapsto\beta_2\mapsto\cdots\mapsto\beta_{\ell-1}\mapsto\beta_0.$$
Consider the polynomial
$$
g(x)=(x-\beta_0)(x-\beta_1)(x-\beta_2)\cdots(x-\beta_{\ell-1})=\sum_{i=0}^\ell g_ix^i.
$$
The usual argument shows that the elements $\phi_p(\beta_i),i=0,1,\ldots,\ell-1,$ are the zeros of the polynomial $(\phi_pg)(x)=\sum_{i=0}^\ell\phi_p(g_i)x^i$. Therefore $g(x)$ and $(\phi_pg)(x)$ are monic polynomials of same degree and same set of pairwise distinct zeros, so they must be equal. Therefore $\phi_p(g_i)=g_i$ for all $i$. Only the elements of the prime field $\Bbb{F}_p$ are fixed under $\phi_p$, so we can conclude that $g(x)\in\Bbb{F}_p[x]$. But the zeros of $g(x)$ are all also zeros $f(x)$, so $g(x)\mid f(x)$. Because $f(x)$ was assumed to be irreducible, we can conclude that $g(x)=f(x)$, $\ell=n$ and that $\phi_p$ permutes the zeros of $f(x)$ in an $n$-cycle.
The case $f=\prod_i f_i$ is now easy. The Frobenius automorphism $\phi_p$ will permute the zeros of a given irreducible factor $f_i$ among themselves. Applying the preceding argument to the irreducible polynomial $f_i$ we see that this permutation is an $n_i$-cycle. Therefore $\phi_p$ permutes the zeros $f_1$ in an $n_1$-cycle, those of $f_2$ in an $n_2$-cycle et cetera. Therefore it has the cycle type $(n_1,n_2,\ldots,n_r)$ WHEN VIEWED AS A PERMUTATION OF THE ZEROS OF $f$.
You knew that in the latter case the splitting field of $f$ is $\Bbb{F}_{p^m}$. If we construct this field as the splitting field of an irreducible polynomial $h(x)$ of degree $m$, and identify $\phi_p$ as permutation of the roots of $h$, then it acts as an $m$-cycle. This is what my rant in the first two paragraphs was about. To get a handle on the cycle structure of $\phi_p$ we need to specify what set it is acting on.
