Let $X$ be a set with $n$ elements and let $G$ be an abelian group acting on $X$ such that: $$(i) \space gx=x \space \forall x \implies g=1,$$ $$(ii) \space \forall x,y \in X, \exists g: gx=y.$$
Show that if $gx=x$ for some $x \in X$, then $g=1$. Deduce that $|G|=n$.
I am not so sure how to attack this problem.
From (i), if $g \in \bigcap_{x \in X} G_x$, then $g=e_G$, so $\bigcap_{x \in X} G_x=\{e_G\}$.
If there is some $x \in X$ such that $gx=x$ and $y \in X$, by (ii) we have that there is some $g'$ such that $g'x=y$, then $gg'x=gy$, using the fact that $G$ is abelian and the group action property "$(gh).x=g.(h.x)$", we obtain $$gy=gg'x=g'gx=g'x=y $$, by (ii), it must be $g=e_G$.
I got stuck here, I would appreciate some help.