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.

Remember that finite fields are unique up to isomorphism. This follows from the uniqueness of splitting fields and the fact that a finite field of order $q=p^k$ is exactly the splitting field of $x^q-x$ (over $\mathbb{F}_p=\mathbb{Z}_p$).
If you had more than one copy of the field of order $q$, you'd have more than $q$ roots for the polynomial $x^q-x$ (contradiction).
So yes your "$K_i$" subfields would all have to be equal. 
In the end, finite fields work a lot like (finite) cyclic groups (which have a unique subgroup for each divisor order).
