Let $G \leq \operatorname{Sym}(\Omega)$ be a permutational group. A base for $G$ is a sequence of elements of $\Omega$ whose pointwise stabilizer is the identity. A base is called irredundant if no point is fixed by the stabilizer of its predecessors. $G$ is said to be IBIS if all of its irredundant bases have the same cardinality. We set $b(G)$ to be the minimal size of a base of $G$.
Currently, I am working on the classification of almost simple IBIS group, and I came up with a particular condition of which I want to ask to you if you have ever seen it.
In particular, I noted that if $G$ is IBIS with $b(G)=3$, then, for $\alpha, \beta \in \Omega$ such that $\beta$ is not fixed by the stabilizer of $\alpha$, then
\begin{equation}
C_G(x) \leq N_G(G_{\alpha} \cap G_{\beta}) \, \forall x \in G_{\alpha} \cap G_{\beta}.
\end{equation}
So I am wondering if it is possible to determine all the couples $(G,H)$, with $G$ an almost simple group, $H \leq G$ core-free, such that
\begin{equation}
C_G(x) \leq N_G(H) \, \, \forall x \in H, \, x \neq 1.
\end{equation}
