Consider first a finite group, $G$, and an involutive automorphism, $s$, such that the automorphism corresponding to $s^2$ is the identity automorphism. Now consider the subgroup $K(s)$ of $G$ such that an element $k$ of $K(s)$ is such that $s(k)=k$ meaning that the involutive automorphism $s$ applied $k$ is fixed (same as identity applied to $k$). Is it possible to prove that $K(s)$ is a normal subgroup of $G$ meaning that for any element $g$ of $G$ and $k$ of $K(s)$ then $g.k.g^{-1}$ is a member of $K(s)$ and if so how would one go about proving it? If it is not true then can anyone give a counter example. If it is any easier take the specific example that the automorphism $s$ is the conjugate transform $s.g.s, s=s^{-1}$. In this case i need to show $s.g.k.g^{-1}.s=g.k.g^{-1}$.
What about the same idea applied to an infinite order group or some Lie group? Now that i think about it i guess for the finite group conjugate transform example a good place to start may be consider the permutation groups starting with the $6$ elements of $S_3$ and see if it is or is not true .