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Re-edited to include Derek Holt's observation: This question seems to depend on the smallest prime divisor $p$ of $n.$ I am not sure how to proceed if $n =p.$ For example, if $n =p$ and every doubly transitive permutation group of degree $p$ is solvable, then $S_p$ has no transitive subgroup of order greater than $p(p-1)$. Using the classification of finite simple groups, it seems likely that there are infinitely many such primes,(but this is not entirely straightforward, since it depends whether or not $p$ can be represented by certain cyclotomic polynomials). If, however, $n$ is not prime, then $p \leq \sqrt{n}$ and $S_n$ has the large transitive subgroup $ S_{\frac{n}{p}}\wr S_{p}$

Re-edited to include Derek Holt's observation: This question seems to depend on the smallest prime divisor $p$ of $n.$ I am not sure how to proceed if $n =p.$ For example, if $n =p$ and (apart from $S_{p}$ and $A_{p}$) every doubly transitive permutation group of degree $p$ is solvable, then $S_p$ has no transitive subgroup of order greater than $p(p-1),$ other than $A_{p}$. Using the classification of finite simple groups, it seems likely that there are infinitely many such primes,(but this is not entirely straightforward, since it depends whether or not $p$ can be represented by certain cyclotomic polynomials). If, however, $n$ is not prime, then $p \leq \sqrt{n}$ and $S_n$ has the large transitive subgroup $ S_{\frac{n}{p}}\wr S_{p}$

This question seems to depend on the smallest prime divisor $p$ of $n.$ I am not sure how to proceed if $n =p.$ For example, if $n =p$ and every doubly transitive permutation group of degree $p$ is solvable, then $S_p$ has no transitive subgroup of order greater than $p(p-1)$. Using the classification of finite simple groups, it seems likely that there are infinitely many such primes,(but this is not entirely straightforward, since it depends whether or not $p$ can be represented by certain cyclotomic polynomials). If, however, $p$ is small relative to $n,$ then $S_n$ has the large transitive subgroup $ S_{p} \wr S_{\frac{n}{p}}.$

Re-edited to include Derek Holt's observation: This question seems to depend on the smallest prime divisor $p$ of $n.$ I am not sure how to proceed if $n =p.$ For example, if $n =p$ and every doubly transitive permutation group of degree $p$ is solvable, then $S_p$ has no transitive subgroup of order greater than $p(p-1)$. Using the classification of finite simple groups, it seems likely that there are infinitely many such primes,(but this is not entirely straightforward, since it depends whether or not $p$ can be represented by certain cyclotomic polynomials). If, however, $n$ is not prime, then $p \leq \sqrt{n}$ and $S_n$ has the large transitive subgroup $ S_{\frac{n}{p}}\wr S_{p}$

This question seems to depend on the smallest prime divisor $p$ of $n.$ I am not sure how to proceed if $n =p.$ For example, if $n =p$ and every doubly transitive permutation group of degree $p$ is solvable, then $S_p$ has no transitive subgroup of order greater than $p(p-1)$. Using the classification of finite simple groups, it seems likely that there are infinitely many such primes,(but this is not entirely straightforward, since it depends whether or not $p$ can be represented by certain cyclotomic polynomials). If, however, $p$ is small relative to $n,$ then $S_n$ has the large transitive subgroup $ \left( \mathbb{Z}/p\mathbb{Z} \right) \wr S_{\frac{n}{p}}.$

This question seems to depend on the smallest prime divisor $p$ of $n.$ I am not sure how to proceed if $n =p.$ For example, if $n =p$ and every doubly transitive permutation group of degree $p$ is solvable, then $S_p$ has no transitive subgroup of order greater than $p(p-1)$. Using the classification of finite simple groups, it seems likely that there are infinitely many such primes,(but this is not entirely straightforward, since it depends whether or not $p$ can be represented by certain cyclotomic polynomials). If, however, $p$ is small relative to $n,$ then $S_n$ has the large transitive subgroup $ S_{p} \wr S_{\frac{n}{p}}.$