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.

This follows from the fact that the algebraic closure of a finite field $\mathbb F_{p^r}$ is simply the union of $\mathbb F_{p^n}$ for all $n$.  Moreover, if $k = \overline{\mathbb F_{p^r}}$ is this algebraic closure then $\mathbb F_{p^n}$ is given exactly as the set of roots in $k$ of the polynomial $x^{p^n} - x$.
This means that in $k$ there is only one subfield of size $p^r$.  A nonzero homomorphism $\mathbb F_{p^r} \to k$ would have as image a field of size $p^r$ so its image would be $\mathbb F_{p^r}$ again.
