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*}
$
 A: 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 $\operatorname{ker}\lambda:=\{g\in G:\lambda(g)=e\}$.
A: In general, if we are given an action of a group $G$ on a set $X$, $G\times X\to X$, we can establish a homomorphism $\lambda\colon G \to \operatorname{Sym}(X)$ by defining $\lambda_g(x):=g\cdot x$. In fact, $\forall g,h\in G,\forall x\in X$:
$$\lambda_{gh}(x)=(gh)\cdot x=g\cdot(h\cdot x)=\lambda_g(\lambda_h(x))=(\lambda_g\lambda_h)(x)$$
whence:
$$\lambda_{gh}=\lambda_g\lambda_h, \space\forall g,h\in G \tag 1$$
(That indeed $\lambda_g \in \operatorname{Sym}(X), \forall g \in G$, also follows from action's axioms.)
The kernel of this homomorphism (the so-called "kernel of the action") is given by:
\begin{alignat}{1}
\operatorname{ker}\lambda &= \{g\in G\mid\lambda_g=\iota_X\} \\
&= \{g\in G\mid\lambda_g(x)=\iota_X(x), \forall x \in X\} \\
&= \{g\in G\mid g\cdot x=x, \forall x \in X\} \\
&= \{g\in G\mid g\in \operatorname{Stab}(x), \forall x \in X\} \\
&= \{g\in \operatorname{Stab}(x), \forall x \in X\} \\
&= \bigcap_{x\in X}\operatorname{Stab}(x) \\
\tag 2
\end{alignat}
In your case, $X=\{gH, g\in G\}$ and $\operatorname{Stab}(gH)=\{g'\in G\mid g'gH=gH\}$.
Lemma. $\operatorname{Stab}(gH)=gHg^{-1}$.
Proof.
\begin{alignat}{1}
g'\in \operatorname{Stab}(gH) &\Rightarrow \exists h,h'\in H\mid g'gh=gh' \\
&\Rightarrow \exists h,h'\in H\mid g'=gh'(gh)^{-1}=gh'h^{-1}g^{-1} \\
&\Rightarrow \exists h''\in H\mid g'=gh''g^{-1} \\
&\Rightarrow g'\in gHg^{-1} \\
\end{alignat}
and thence $\operatorname{Stab}(gH)\subseteq gHg^{-1}$. Viceversa:
\begin{alignat}{1}
g'\in gHg^{-1} &\Rightarrow g'g \in gH \\
&\Rightarrow g'gH\subseteq gH \\
\tag 3
\end{alignat}
Now:
\begin{alignat}{1}
gH\subseteq g'gH &\iff \forall h \in H, \exists h'\in H\mid gh=g'gh' \\
&\iff \forall h \in H, \exists h'\in H\mid gh=(gh''g^{-1})gh'=gh''h' \\
&\iff h'=(gh'')^{-1}gh=h''^{-1}g^{-1}gh=h''^{-1}h \\
\end{alignat}
Since $h''$ exists by hypothesis ("Viceversa..."), such a $h'$ exists, and then indeed:
$$gH\subseteq g'gH \tag 4$$
Therefore, $g'\in gHg^{-1} \Rightarrow g'gH=gH \Rightarrow g'\in \operatorname{Stab}(gH) \Rightarrow gHg^{-1}\subseteq \operatorname{Stab}(gH)$.
$\Box$
By $(2)$ and the Lemma, we have finally:
$$\operatorname{\ker}\lambda = \bigcap_{gH\in G/H}\operatorname{Stab}(gH)= \bigcap_{g\in G}\operatorname{Stab}(gH)=\bigcap_{g\in G}gHg^{-1} \tag 5$$
