Are these two inclusions of finite groups, equivalent? Definition :  Let $\sim$ be the equivalence relation on inclusions of finite groups, generated by :
$(H \subset G) \sim (\phi(H) \subset \phi(G))$, with $ \phi: G \to L$  a group morphism, and $ker(\phi) \subset H$.   

Question :  $(\langle (1234)   \rangle \subset S_4 ) \sim (\langle
 (12),(34)   \rangle \subset S_4 )$ ?    

Warning: Of course, there is no $\phi \in Aut(S_4)$ with $\phi(\langle (1234)   \rangle) = \langle (12),(34)   \rangle$,   because $\langle (1234)   \rangle \simeq \mathbb{Z}_4 \not\simeq \mathbb{Z}^2_2 \simeq \langle (12),(34)   \rangle$, that why $\sim$ is the equivalence relation generated by the relation above, not just the relation alone.    
Remark : the non-trivial normal subgroups of $S_4$ are $A_4$ and $\langle
 (12)(34), (13)(24)   \rangle$ (see here).
Motivation : These inclusions are famous in the group-subgroup subfactor theory (see here p47).
 A: I'm not an expert in finite groups theory, so I will try to develop the comment of Jack Schmidt :  
Let $(H \subset G)$ and $\phi:G \to L$ with $ker(\phi) \subset H$.
Let $\Omega=G/H$ and $\Omega' = \phi(G) / \phi(H)$.   
Let the map $\delta : \Omega \to \Omega'$ : $\delta(gH) = \phi(g)\phi(H)$
$\delta$ is well-defined because if $g_1^{-1}g_2 \in H$ then $\phi(g_1)^{-1}\phi(g_2) = \phi(g_1^{-1}g_2) \in \phi(H)$
$\delta$ is then obviously surjective, and it's injective because if $\phi(g_1)^{-1}\phi(g_2) \in \phi(H)$ then $g_1^{-1}g_2 \in H$, because for $h \in H$, $\phi^{-1} (\{\phi (h) \}) = h.ker(\phi) \subset H$.  
Let $\pi: G \to S_{\Omega} : \pi(g_1)(g_2H) = g_1g_2H$, then $\pi(G) \simeq G/ker(\pi) $
$ker(\pi) = \{ g_1 \in G: g_1g_2H = g_2H,  \forall g_2 \in G \} = \bigcap_{g_2 \in G} g_2^{-1}Hg_2 = core_{G}(H) \subset H$ is the normal core of $H$ in $G$ i.e the largest normal subgroup of $G$ contained in $H$.
Idem, let $\pi': \phi(G) \to S_{\Omega'} $, then  $\pi'(\phi(G)) \simeq \phi(G)/ker(\pi') $  
Now $\pi(G)$ and $\pi'(\phi(G))$ are permutation isomorphic because $\delta$ is a bijection and obviously $\delta(\pi(g_1)(g_2H)) = \pi'(\phi(g_1))(\phi(g_2)\phi(H))$;  in particular $\pi(G) \simeq \pi'(\phi(G))$.  
Conclusion, the equivalence relation generated by:
$$(H \subset G) \sim (\phi(H) \subset \phi(G)) \text{ with }   \phi: G \to L \text{ morphism  and } ker(\phi) \subset H$$  implies this equivalence relation  $\sim'$ as above (equivalence of permutation actions on the cosets).    
Optional question : What's about the converse ?   

Anyway, about my example :
If $(\langle (1234)   \rangle \subset S_4 ) \sim (\langle
 (12),(34)   \rangle \subset S_4 )$, then $(\langle (1234)   \rangle \subset S_4 ) \sim' (\langle
 (12),(34)   \rangle \subset S_4 )$ and so $\langle (1234)   \rangle \simeq \langle
 (12),(34)   \rangle$, because the normal cores are trival,  contradiction.      
The normal cores are trival because there is not non-trivial normal subgroup of $S_4$ contained in $\langle (1234)   \rangle $ or $ \langle
 (12),(34)   \rangle$, thanks to the classification of the normal subgroups of $S_4$ given above.
