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Modulo three, the quintic has no linear factor: its value is $-1$ at every element of ${\mathbb{F}}_3$. The monic irreducible quadratics over this field are $t^2+1$, $t^2+t+2$ and $t^2+2t+2$. Division leaves respectively $-1$, $t-1$ and $t-1$, all nonzero. A reducible quintic has a factor of degree at most two, so it is irreducible modulo three and therefore over $\mathbb{Q}$. Its Galois group is transitive on five roots.

Modulo two its squarefree factorization isBoth factors are irreducible, so the Frobenius cycle type theorem gives an element that is a disjoint transposition times a three-cycle. Cubing it gives a transposition.

The prime-degree transposition criterion can be proved here: a transitive subgroup of prime degree containing a transposition is the whole symmetric group. Construct a graph on the five roots whose edges are the pairs swapped by all conjugates of this transposition. The group acts transitively on vertices and permutes connected components, so all component sizes are equal and divide five. An edge rules out size one; the graph is connected. Edge transpositions of any connected graph generate the full symmetric group, for instance by transporting swaps along a spanning tree. All belong to the Galois group, provingThe prime-degree condition is essential to the short transposition argument.

The quartic is irreducible by Eisenstein criterion at two, so its Galois group is transitive on its four roots. Its discriminant is $1616=16\cdot 101$, so five and seven are unramified primes for the polynomial. The squarefree factorizations areThe cubic modulo five has no root in ${\mathbb{F}}_5$, hence is irreducible. The quadratic modulo seven has discriminant five, a nonsquare there. The Frobenius cycle type theorem therefore supplies a three-cycle and a transposition in the group.

Transitivity implies that its order is divisible by four, and the three-cycle implies divisibility by three. Its order divides $24$, so it is $12$ or $24$. The only subgroup of index two in $S_4$ is $A_4$: an index-two subgroup is the kernel of a group homomorphism of a nontrivial homomorphism to $C_2$, and all transpositions are conjugate and generate $S_4$, forcing this homomorphism to be the sign. The exhibited transposition excludes $A_4$. Hence

Orbits of the root action are precisely roots belonging to the same irreducible factor, proving transitivity iff irreducibility.

For a same-degree counterexample, let $L=\mathbb{Q}(\sqrt{2},\sqrt{3})$,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 $S_4$.