Let $H$ be a subgroup of $G$, and suppose that $G$ acts by multiplication over the set $X:=G/H$ of the left-hand side classes of $H$ over $G$.

This action is transitive. Now let be $yH \in X$. What is the kernel of this action?

I'm afraid that my answer isn't right. Could you check if the KERNEL is exactly this?

MY ANSWER:

Let $A:G \times X \rightarrow [G:H]=X$ be the action mentioned and $\lambda:G\rightarrow \operatorname{Sym}(G)$ (this homomorphism exists because of the permutational representation). So, we have

$ \begin{align*} \operatorname{ker}(A)&=\{(g,yH) \in A: A(g, yH)=gyH=yH, \forall (g, yH)\}\\ &=\{g \in G: gyH=yH, \forall g \in G\}\\ &=yHy^{-1}, \forall g \in G\; \text{the reason of this step is going to be explained below**}\\ &=\displaystyle\bigcap_{g \in G} yHy^{-1} \end{align*} $

Explanation about **:

$ \begin{align*} \operatorname{Stab}(yH)&=\{g \in G: gyH=yH\}\\ &=\{g \in G: y^{-1} gyH=H\}\\ &=\{g \in G: y^{-1}gy \in H\}\\ &=\{g \in G: g \in yHy^{-1}\}\\ &=yHy^{-1} \end{align*} $

The kernel of $\lambda$ should be $\bigcap_{g\in G}gHg^{-1}$. This is called the normal core of $H$.

I think this is what you want. $A$ doesn't have a kernel, as $X$ is only a set. But $\operatorname{Sym}X$ is a group, and $\lambda$ a homomorphism. So we can talk about $\operatorname{ker}\lambda:=\{g\in G:\lambda(g)=e\}$.