Let $K$ be the splitting field of a separable irreducible polynomial $f(x) \in F[x]$ of degree $n$ and let $G = Gal(K/F)$. If for each $g \in G$, there is a root $\alpha$ of $f$ such that $g(\alpha) = \alpha$, prove that $K=F$.
We know that $G$ permutes the roots of $f$. Since $f$ is irreducible separable over $F$, $G$ is isomorphic to a transitive subgroup of $S_n$, i.e., $G$ has some $g$ that is an $n$-cycle. But at the same time, $g$ also fixes some root of $f$ and so $n=1$.
Is that all? I've thought long and hard about this only to come up with a literal one-liner.
Suppose $K\neq F$.
Let $\{\alpha_1,\ldots,\alpha_n\}$ be the distinct roots of mimimal pol. of $\alpha$ over $F$ ($\alpha=\alpha_1$), and $G=\{\sigma_1,\ldots,\sigma_n\}$.
Take any $1\neq \sigma\in\mbox{Gal}(K/F)$. Since $\sigma$ fixes some $\alpha_i$, so $\mbox{Stab}(\alpha_i)$ is non-trivial subgroup.
Since $\alpha_i\notin F$, so $\alpha_i$ is not fixed by $\mbox{Gal}(K/F)$, so $\mbox{Stab}(\alpha_i)$ is proper subgroup.
For any other $\alpha_j$,there is $\tau\in\mbox{Gal}(K/F)$ such that $\tau(\alpha_i)=\alpha_j$. Then $$ \tau\mbox{Stab}(\alpha_i)\tau^{-1}=\mbox{Stab}(\tau(\alpha_i))=\mbox{Stab}(\alpha_j). $$ Thus, the given condition says: every $\sigma\in\mbox{Gal}(K/F)$ is in some conjugate of $\mbox{Stab}(\alpha_i)$, i.e. $$ \mbox{Gal}(K/F)=\cup_{\tau} \tau \mbox{Stab}(\alpha_i)\tau^{-1}. $$ But, a finite group can never be union of conjugates of a proper subgroup. Q.E.D.