Does the kernel need to be the full automorphism group of the induced subgraph? Let $\Gamma$ be a simple graph. Suppose its automorphism group $G=\text{Aut}\Gamma$ is imprimitive on its vertex set $V$. Take a block system $\mathcal{B}$ of $V$. Let $B\in\mathcal B$, and let $K=G_{(\mathcal B)}$. If $K$ acts faithfully on the induced subgraph $[B]_\Gamma$, that is $K_{(B)}=1$, can we always say $K^B=\text{Aut}[B]_\Gamma$? 

NB: $G_{(\mathcal B)}$ is the subgroup of $G$ which stabilises $\mathcal B$ pointwisely. 
$G^\mathcal{B}:=G/G_{(\mathcal B)}$ is the group that acts on $\mathcal B$ faithfully.
The same way of definition applies to all other notations wherever it applies.
 A: The answer is "no"; I offer one counterexample.
I assume that imprimitive means vertex transitive and imprimitive (an imprimitive group
need not be transitive).
Take $X$ to be the icosahedron, a 5-regular arc-transitive
graph on 12 vertices with automorphism group $G$ of order 120.
Now let $Y$ be the line graph of the subdivision graph of the icosahedron, i.e., 
put a new vertex in the middle of each of the 60 edges of the icosahedron, then take the line graph of this subdivided graph. 
Then $|V(Y)|=2|E(X)|=120$ and $H=\mathrm{Aut}(Y)$ is isomorphic to $G$ (by
theorems about automorphism groups of line graphs).
Since $G$ is arc transitive, $H$ is vertex transitive, in fact it acts regularly on the vertices of $Y$. Each vertex of $X$ gives rise to a clique 
of size five in $Y$, and these cliques are blocks of imprimitivity for $H$. The subgroup of $H$ that fixes one of these cliques has order at most five, and so it does not induce the full automorphism group of the clique.
I am sure this construction will work with many other arc-transitive graphs.
