Is the following scenario possible for stabilizer

Question

So I was solving couple of problems involving stabilizer of group actions. I was wondering can we have kernel of group action being equal to the identity while all the stabilizers of G not being equal to the identity ? I mean I know such scenrio isn't possible if the group if we consider abelian subgroup of $S_A$ with regular action on elements of A,however I don't know what would happen in general case. It seems that such scenrio would be possible even if we think of it in terms of set theoretic terms. That is the kernel of the action being the intersection of all of the stabilizers.

Answer 1

Sure. The usual action of $S_n$ on the set $\{1,2,\ldots,n\}$ has this property. The kernel of this action is trivial, since the action is faithful. However, all point stabilizers are canonically isomorphic to $S_{n-1}$.

Answer 2

Take the free group of rank $2$ acting on itself by conjugation. Its center is trivial so the action has a trivial kernel. However every element has a non-trivial centralizer hence no stabilizer can be trivial.