In the discussion below we are assuming that $F$ is a field of characterisc $0$.
Let $K$ and $K'$ be finite extensions of the field $F$ and $K$ and $K'$ have a commom extension $E$. If $h:K \rightarrow K'$ is a isomorphism such that $h(x) =x$ for every $x$ in $F$ and $K$ is the root field of some polynomial $a(x)$ over $F$ then $h$ maps elements of $K$ to elements of $K$.
The problem begins when Pinter states the following theorem.
Theorem: Let $K$ and $K'$ be finite extensions of $F$. Assume $K$ is the root field of some polynomial over $F$. If $h: K\rightarrow K'$ is a isomorphism which fixes $F$, then $K = K'$.
Proof: We know that $K$ and $K'$ are simple extensions of $F$, say $K=F(a)$ and $K'=F(b)$. Then $E=F(a,b)$ is a commom extension of $K$ and $K'$. By the comments preeceding this theorem, $h$ maps every element of $K$ to an element of $K'$; hence $K' \subseteq K$. Since the same argument may be carried out for $h^{-1}$, we also have $K \subseteq K'$. QED
I can understand the reason why $K' \subseteq K$, but I don't know how to proceed to prove the other inclusion ( $K \subseteq K'$), if someone can help me understand this theorem, I will really appreciate. For futher details, this situation can be found on chapter 31 page 316 of the second edtion of Pinter's book.