If $G$ is non-abelian group of order 6, it is isomorphic to $S_3$ Let $G$ be a non-abelian group of order $6$ with exactly three elements of order $2$. Show that the conjugation action on the set of elements of order $2$ induces an isomorphism. 
I just need to show that the kernel of the action is trivial. Not sure how to go about doing that. I think maybe a proof by contradiction but I can't find a contradiction. I would think it would violate "non-abelian-ness" of the group. Thanks for any help!

Hint: Suppose $x\in G$ is an element of the kernel of the action, i.e. fixes the three involutions under conjugation. What do you know about the group generated by the three involution, and what does that tell you about $x$?
