= Solution
Orbits of the root action are precisely roots belonging to the same irreducible factor, proving transitivity iff irreducibility.
For a same-degree counterexample, let $L=\mathbb Q(\sqrt2,\sqrt3)$,
$$
f=(X^2-2)(X^2-3),\qquad
g=\min_{\mathbb Q}(\sqrt2+\sqrt3)=X^4-10X^2+1.
$$
Both split in $L$, but the first action has two orbits of size two while the second is transitive, so the <subgroups> are not conjugate in $S_4$.
Over a finite field, <Frobenius cycle type> proves the assertion. The given quartic reduces over $\mathbb F_2$ to $X^4+X^3+1$, which is irreducible, so its <group> there is cyclic of order four generated by a $4$-cycle.
Over $\mathbb Q$ the <polynomial> is irreducible by the same reduction. Modulo $3$ it factors as linear times irreducible cubic, so <Dedekind factorization cycle test> gives a $3$-cycle; modulo $2$ it gives a $4$-cycle. Its discriminant is $229$, and modulo $23$ its factor degrees are $1,1,2$, giving a transposition. These cycle types force the transitive <group> to be $S_4$.
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