Let $M$ be a homogeneous space, i.e. there is a group $G$ acting transitively on $M$. Let $p$ be a point in $M$ with stabiliser $H$ in $G$. Then $M$ is isomorphic to $G/H$ via the equivariant correspondence $gH \in G/H \mapsto gp \in M$.
Now write $p = gH$ for some $g \in G$. If you take $h \in H$ then $h p = p$, which by the equivariance above means $hgH = gH$, or equivalently $g^{-1}hg \in H$. But this holds for all $h \in H$, thus $g$ sits in the normaliser $N_G(H)$ of $H$ in $G$. Therefore $p \in N_G(H)/H$. But this shows $G/H \subset N_G(H)/H$ via the trivial immersion, thus $N_G(H) = G$, which is not true in general.
Where is the flaw in my argument?
So lets write everything down in a bit more precise way. We start with a group $G$ and a (transitive) $G$-set $M$. We pick a point $p\in M$ and we consider its stabilizer group $H_p\subseteq G$. So the first important thing is that $H_p$ is tightly related to $p$ (hence the index). And yes, for different $p$ these subgroups are always conjugate, but it isn't really relevant. Then as you've already said there's a $G$-isomorphism
$$f_p:G/H_p\to M$$ $$f_p(gH_p)=g\cdot p$$
And here things become blurry. You say "Now write $p=gH$". There is no equality! There is an isomorphism only. And this is a crutial observation. First of all if you write $f_p^{-1}(p)=gH_p$ then in fact this is only true when $g\in H_p$ to begin with.
But then, you may say, this is even worse because that stronger claim (i.e. $g\in H_p$) implies that $G/H\subseteq H/H$ by the arbitrary choice of $p$, right? No. If we change $p$ then our $H_p$ subgroup may change as well. But even if it doesn't change (say $H_p$ is normal), our $f_p$ isomorphism definitely changes. That's why regardless of the choice of $p$ we always have $f_p^{-1}(p)=H_p$! We never leave $H_p\in G/H_p$ coset, regardless of $p$.
The moral of this story is: be very cautious when treating isomorphisms like equality. It does not always work.