Let $G=X$. Is the following a group action?
$$g(x)=gxg$$
If it is, what's the stablizer?
I had troubles with the second group action axiom. I wasn't sure if I did correctly. Also, on a side note, if the group action is on itself, isn't the first axiom always correct by the associativity of a group?
As pointed out by Alex G. in the comments, it is not a group action as $(gh)\cdot x = ghxgh$ whereas $g\cdot(h\cdot x) = g\cdot (hxh) = ghxhg$.
As for your second question, the answer is no. Consider the map $\mathbb{Z}\times\mathbb{Z} \to \mathbb{Z}$ given by $a\cdot n = an$. This does not satisfy the first group action axiom despite the fact that the group is 'acting' on itself.