Let $n\ge 3$ and $X=\{1,2,\cdots,n\}$ and $P=(P_1,P_2,\cdots,P_s)$ a partition of $X$
Define $S_n$ as the set of permuations of $X$
Let $S_{n,P}:=\{\sigma \in S_n| ~~\sigma(P_i)=P_i ~~ \forall 1\le i\le s\}$
Find a sufficient and necessary condition on $P$ such that $S_{n,P}$ is a maximal subgroup of $S_n$ (it is clearly a subgroup of $S_n$)
Progress: I proved that $s=2$ because otherwise we can just take a transposition $(a~~ b)$ with $a\in P_1$ and $b\in P_2$ and add it to $S_{n,P}$ and get a group which has $S_{n,P}$ in it and is different from $S_n$ because $P_3$ is still stable by any permutation from that group
Now i guess I have to prove that $|P_1|\neq |P_2|$ and prove that in the other cases $S_{n,P}$ is maximal. I ve proved for $|P_2|=1$ but I haven't done the rest yet.
Here is a sketch proof. You have proved that $s=2$, so $S_{n,P} = S_{n_1} \times S_{n_2}$, where $n_i = |P_i|$ for $i=1,2$.
If $n_1 = n_2$ then $S_{n,P}$ has index $2$ in $S_{n_1} \wr C_2$, which consists of those permutation which either fix both sets $P_i$ or interchange them, so it is not maximal.
So we can assume that $n_1 > n_2$. We have to prove that $S_{n,P} < K \le S_n \Rightarrow K = S_n$, so suppose that $S_{n,P} < K \le S_n$. Then there exists $g \in K \setminus S_{n,P}$.
Then $g$ must map some point $\alpha \in P_1$ to a point $\alpha' \in P_2$. But, since $n_1>n_2$, it cannot map all points in $P_1$ to points in $P_2$, so it maps some $\beta \in P_1$ to $\beta' \in P_1$.
Now $g^{-1} (\alpha,\beta) g = (\alpha',\beta')$ (I compose permutations left to right), so $(\alpha',\beta') \in K$.
But now, for any transpositions $(\gamma,\delta)$ with $\gamma \in P_2$, $\delta \in P_1$, there exists $h \in S_{n,P}$ that maps $\alpha'$ to $\gamma$ and $\beta'$ to $\delta$ and so $(\gamma,\delta) =h^{-1}(\alpha',\beta')h \in K$.
All other transpositions in $S_n$ lie in $S_{n,P}$, so So $K$ contains all transpositions, and hence $K=S_n$.