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Well I think I have more or less an answer to my question. I have shown that the set of all maximal imprimitive transitive subgroups $H\leq S_N$ is of the form $$ S_{N/r}^{r}\rtimes S_r $$ for $r|N$ and where $S_r$ acts by permutation on the coordinates of $S_{N/r}^r$. So since I have an onto group homomorphism $$ f:H\rightarrow S_n $$ I must conclude that $H\subseteq S_{2}^{n}\rtimes S_n$ and that $H\supseteq S_n$. Finally, since I can produce an element $\tau\in H$ that has a cycle of length larger than $n$ which appears in its cycle presentation I may conclude that $H$ is not contained in any maximal transitive imprimitive subgroups of $S_N$ and therefore by maximality this implies that $H=S_N$. But this is absurd since it contradicts the imprimitivity of $H$. Therefore such an $H$ does not exist.