Transitive subgroups of $S_4$ are $S_4$, $A_4$, $D_4$, $\ZZ_2^2$, and $\ZZ_4$

Let \(G\le S_4\) be transitive. Orbit-stabilizer gives \(4\mid |G|\), while Lagrange gives \(|G|\mid24\). Hence \[ |G|\in\{4,8,12,24\}. \] If \(|G|=24\), then \(G=S_4\). If \(|G|=12\), then \(G\) has index \(2\), hence is the kernel of a nontrivial homomorphism \(S_4\to C_2\); since \(S_4^{\mathrm{ab}}\cong C_2\), this subgroup is \(A_4\).

If \(|G|=8\), then \(G\) is a Sylow \(2\)-subgroup of \(S_4\). All Sylow \(2\)-subgroups are conjugate, and one is the symmetry group of a square on the four letters, hence isomorphic to \(D_8\) (dihedral of order \(8\)).

If \(|G|=4\), transitivity makes the action regular. The two groups of order \(4\) both occur regularly: \(C_4=\langle(1234)\rangle\) and \[ V_4=\{1,(12)(34),(13)(24),(14)(23)\}. \] Therefore the transitive subgroups of \(S_4\), up to conjugacy/isomorphism as appropriate, are \[ S_4,\quad A_4,\quad D_8,\quad C_4,\quad V_4. \]