How might I calculate $\text{Der}_{\mathbb{F}_p} ( \mathbb{F}_{p^n}, \mathbb{F}_p^{sep} )$, given a choice of embedding $\mathbb{F}_{p^n} \rightarrow \mathbb{F}_p^{sep}$? Further, what is $\text{Der}_{\mathbb{F}_p} ( \mathbb{F}_{p}^{sep}, \mathbb{F}_p^{sep} )$?
This is like the tangent space of $\text{Spec}(\mathbb{F}_{p^n})$ at one of the $\mathbb{F}_p^{sep}$-points, in correspondence with $\mathbb{Z}/n \mathbb{Z}$, which has an action of the absolute galois group $\mathbb{F}_p^{sep}$.
Suppose that $k$ is a commutative ring, $A$ is an asociative, commutative $k$-algebra, and $M$ is an $A$-module. If $\phi:A\to M$ is a derivation, then induction shows that $\phi(x^n)=nx^{n-1}\phi(x)$, and so for any polynomial $f$, $\phi(f(x))=f'(x)\phi(x)$.
If $A$ is cyclic, so $A=k[x]/I$, then a derivation is completely determined by the image of $x$, and there is a derivation sending $x$ to $m$ so long as $f'(x)m=0$ for every $f(x)\in I$. If $I=(f(x))$, then it suffices that $f'(x)m=0$.
In our case, $k=\mathbb F_p$, $I$ is generated by an irreducible divisor $f(x)$ of $x^{p^n}-x$ of degree $n$, and every nonzero element of $\mathbb F_p^{sep}$ has annihilator $0$, so unless $f'(x)=0$, $x$ must map to $0$.
Can $f'(x)=0$?