Let $H \leq S_N$ be a transitive permutation group. A system of blocks $\Sigma = \{\Delta\}$ for $H$ is a partition of $[N]$ such that either $\Delta^h = \Delta$ or $\Delta \cap \Delta^h = \emptyset$. For primitive $H$, the only system of blocks are the trivial ones $\Delta = \{e\}$ and $\Delta = [N]$. It is also known that there is a one-to-one mapping between the system of blocks of $H$ and overgroups of the stabilizer $H_{n}$ for $n \in [N]$.
My question is this: what can we say about the system of blocks for a larger group $G > H$, $G \leq S_N$? We know the relationship between the stabilizers, that is $H_n < G_n$. I assume the blocks are generally not preserved (?) what more can we say?
There's not a lot to say here, but we can say a couple of things. We'll assume $H\le G\le S_n$ and $H$ and $G$ are both transitive. A block system is defined by a choice of a single block $\Delta\ni 1$, so let $B_H$ be the set of blocks for $H$ containing $1$ and $B_G$ be the set of blocks for $G$ containing $1$.
Immediately if $\Delta^g=\Delta$ or $\Delta^g\cap\Delta=\emptyset$ for all $g\in G$ then certainly this holds for all $g\in H$. So a block for $G$ is a block for $H$ which we can write as $B_G\le B_H$.
The inverse equality clearly doesn't always hold (e.g. $H$ any imprimitive group and $G=S_n$). One can show however that blocks of $G$ containing $1$ are in one-to-one correspondence with subgroups of $G$ containing the point stabiliser $G_1$ (specifically $\Delta$ corresponds to $G_\Delta$ the setwise stabiliser of $\Delta$). I claim that a block $\Delta$ of $H$ containing $1$ is a block for $G$ if and only if $G_\Delta\ge G_1$. This means $B_G=\{\Delta\in B_H|G_1\le G_\Delta\}$ - that is a block of $G$ containing $1$ is any block of $H$ containing $1$ fixed by $G_1$.
Proof of Claim:
"Prove $\Delta$ is a block of $G$ implies $G_1\le G_\Delta$" is a standard exercise, so I will only do the reverse implication.
Assume $G_\Delta\ge G_1$ where $\Delta$ is a block for $H$.
Let $g\in G$ and suppose $x\in\Delta\cap\Delta^g$.
Relabelling if necessary we may assume $x=1$.
As $1\in\Delta^g$ there is some $y\in\Delta$ with $y^g=1$.
As $H$ is transitive there is some $h\in H$ with $1^h=y$ so $1^{hg}=1$.
That is $hg\le G_1\le G_\Delta$.
As $\Delta$ is a block for $H$, $h\le H_\Delta\le G_\Delta$.
Hence $g=h^{-1}(hg)\le G_\Delta$, so $\Delta^g=\Delta$ completing the proof.