Suppose $\mathbb{K}$ is a field and $f(x) \in \mathbb{K}[x]$ is separable. Consider $\mathbb{E}$ is an intermediate field between $\mathbb{F}$, a splitting field of $f$ over $\mathbb{K}$, and $\mathbb{K}$. Is it true that $f$ is irreducible over $\mathbb{E}$ if and only if $Aut_{\mathbb{E}}\mathbb{F}$ is isomorphic to a transitive subgroup of $S_n\ ; n=deg(f)$? (Here $Aut_{\mathbb{E}}\mathbb{F}$ means the set of all field automorphism $\mathbb{F} \rightarrow \mathbb{F}$ which fixes $\mathbb{E}$ elementwise.)
My proof:
Let's say $Aut_{\mathbb{E}}\mathbb{F}$ is transitive if it is isomorphic to a transitive subgroup. Consider $ \{ u_1,\cdots,u_n \} $ to be roots of $f$ in $\mathbb{F}$.
Suppose $f$ is reducible over $\mathbb{E}$, then $f=gh$ ($deg(h) \ge 1$) is decomposed over $\mathbb{E}$. Now, for all $\sigma \in Aut_{\mathbb{E}}\mathbb{F}$, $\sigma$ fixes $g$. Hence, there is no map in $ Aut_{\mathbb{E}}\mathbb{F}$ which sends a root of $g$ to a root of $h$. Hence, $Aut_{\mathbb{E}}\mathbb{F}$ is not transitive.
For the reverse, if $f$ is irreducible over $\mathbb{E}$, there is an $\mathbb{E}$-isomorphism $\sigma: \mathbb{E}(u_i) \rightarrow \mathbb{E}(u_j)$ such that $\sigma(u_i)=u_j$ for every $i,j$. Moreover, $\mathbb{F}$ be a splitting field of $f$ over $\mathbb{K}$ implies $\mathbb{F}$ be a splitting field of $f$ over $\mathbb{E}$. Hence, $\sigma$ can be extended to $\mathbb{F}$, that is $\sigma \in Aut_{\mathbb{E}}\mathbb{F}$. Hence, $Aut_{\mathbb{E}}\mathbb{F}$ is transitive.
Is everything correct?
You have to be a little more careful. As it is stated, the result is actually false. For example, let $K=E=\mathbb Q$ and $f=(X^2-2)(X^2-3)$. Then $F=\mathbb Q(\sqrt2,\sqrt3)$ and the Galois group is $V=(\mathbb Z/2\mathbb Z)^2$. Of course, this is isomorphic to a non-transitive subgroup of $S_4$ using the action on the roots of $f$, so having generators $(12)$ and $(34)$.
However, the Galois group is also isomorphic to a transitive subgroup of $S_4$, using the generators $(12)(34)$ and $(13)(24)$. We can regard this as the action on the conjugates of the primitive element $\sqrt2+\sqrt3$.
In general, every finite Galois extension has a primitive element, and hence the Galois group is always isomorphic to a transitive subgroup of a symmetric group.