Automorphisms on a field $F$ I am trying to understand this proposition with respect to algebraic closures of a field $F$
Prop: If $F$ is a finite field, then every isomorphism mapping $F$ onto a subfield of an algebraic closure $\bar{F}$ of $F$ is an automorphism of $F$
Does this mean that if we have some subfields $K_i \leq \bar{F}$ (for some $i\geq 1$) such that $\exists \Phi: F \rightarrow K_i$ an isomorphism then in reality all these $K_i 's = F$ and the $\Phi's$ are  automorphisms. 
Also how do we go about proving this? Thank you.

Yes, it does mean that for every field homomorphism $\Phi \colon F \to \overline{F}$ we have $\operatorname{im}\Phi = F$.
Every $x \in F$ satisfies the relation $x^n = x$, where $n$ is the number of elements of $F$, thus is a zero of the polynomial $P(X) = X^n - X$. The degree of $P$ is $n$, so $P$ has exactly $n$ zeros in $\overline{F}$ (counting multiplicities, although here they are all $1$).
But if $\Phi \colon F\to \overline{F}$ is a field homomorphism, we have
$$\Phi(x)^n - \Phi(x) = \Phi(x^n-x) = \Phi(0) = 0$$
for every $x\in F$, so $\Phi(x)$ is a zero of $P$, hence an element of $F$. Since field homomorphisms are always injective, and $F$ is finite, $\Phi(F) \subset F$ implies $\Phi(F) = F$.
