Let $p$ be a prime number. Let $a,b,c$ be positive integers, $a$ divides $b$ and $b$ divides $c$.
Fix homomorphisms $f\colon\mathbb{F}_{p^{a}}\to\mathbb{F}_{p^{b}}$ and $g\colon\mathbb{F}_{p^{a}}\to\mathbb{F}_{p^{c}}$.
I want to show that there is a field homomorphism $h\colon\mathbb{F}_{p^{b}}\to\mathbb{F}_{p^{c}}$ such that $h\circ f=g$.
I don't want to use the algebraic closure of $\mathbb{F}_p$ (my question arises from trying to understand the construction of this algebraic closure via direct limits).
Let $K,L,M$ be three finite fields with two homomorphisms
$$\require{AMScd} \begin{CD} K @>{\alpha}>> M \hookrightarrow L \\ @V{\beta}VV \\ L \end{CD}$$
We need to find a homomorphism $\gamma : M \to L$ that makes the triangle commutative. It suffices to find a homomorphism $L \to L$ with that property (because this can then be restricted to $L$). So the task simplifies to completing diagrams of the form
$$\require{AMScd} \begin{CD} K @>{\alpha}>> L \\ @V{\beta}VV \\ L \end{CD}$$ to a homomorphism $\gamma : L \to L$ with $\gamma \circ \alpha = \beta$. In fact, $\gamma$ will be an automorphism.
Notice that $\mathrm{im}(\alpha)$ and $\mathrm{im}(\beta)$ are two subfields of $L$ with the same order. Hence, they are equal. In fact, if $K$ has order $q$ and $L$ has order $q^m$, then $L$ has a unique subfield of order $q$, namely $\{x \in L : x^q = x\}$.
It follows that there is some automorphism $\pi : K \to K$ with $\beta = \alpha \circ \pi$ (we define $\pi$ by $\beta(x) = \alpha(\pi(x))$, it exists since $\beta(x) \in \mathrm{im}(\alpha)$ etc.). So we can completely remove $\beta$ from the task, we only need to find $\gamma$ with $\gamma \circ \alpha = \alpha \circ \pi$.
$$\require{AMScd} \begin{CD} K @>{\alpha}>> L \\ @V{\pi}VV @VV{\gamma}V \\ K @>>{\alpha}> L \end{CD}$$
This can be proven in at least two ways. One proof uses that $\alpha : K \to L$ is a normal extension. Another proof just uses that we actually know all the automorphisms of $K$, they are all powers of the Frobenius $x \mapsto x^p$, where $p = \mathrm{char}(K)$. On $L$, we can then take the same power of the Frobenius, and that's it.