Let $G$ be a finite group acting on a set $A$.
If $b \in \mathcal{O}_a$ i.e ($b=g.a$) for some $g \in G$. Then $G_b=g.G_a.g^{-1}$.
Let $b,c \in \mathcal{O}_a$. If $b\ne c$ then $G_b\ne G_c$
These means that for every element in the orbit, there exists a distinct conjugate of $G_a$. Which means that
(number of conjugates of $G_a$ in $G$) $\geq$ (number of elements in $\mathcal{O}_a$).
Now the LHS=$\frac{|G|}{|N(G_a)|}$ and RHS=$\frac{|G|}{|G_a|}$ $\implies$ $\frac{|G|}{|N(G_a)|} \geq \frac{|G|}{|G_a|}$.
Now since $G$ is finite. Can we conclude that $|G_a| \geq |N(G_a)|$ ?
And since we already know that $|G_a| \leq |N(G_a)|$. We conclude that $N(G_a)=G_a$.
If this is true, then in group actions from any finite abelian group to any set will have trivial stabilizers.
Are these results actually correct?
Note that 2. is not true. You may consider the action from $\mathbb Z/3\mathbb Z$ to an equilateral triangle $ABC$ given by rotations. You have only one orbit, and each vertex has trivial stabilizer.
For a counterexample for your last statement (and also for 2.), you may consider the action of $\mathbb Z$ to an equilateral triangle $ABC$ given by rotations. Then stabilizer of each vertex is $3\mathbb Z$. Even easier, you can take any group $G$ acting trivially (i.e. by formula $g\cdot x=x$) on any set $X$; all stabilizers are equal to $G$, so non-trivial.