Infinite group has no non-identity finitary automorphism

Statement
Let $$G$$ be an infinite group and $$\sigma$$ be an automorphism of $$G$$ that is a finitary permutation of $$G$$: the set of points of $$G$$ that are moved by $$\sigma$$ is finite. Then, $$\sigma$$ is the identity automorphism of $$G$$.

Facts used

 * 1) The set of fixed points under any automorphism $$\sigma$$ of $$G$$ is a subgroup of $$G$$. This subgroup is often termed the centralizer of $$\sigma$$, and denoted $$C_G(\sigma)$$.
 * 2) uses::Proper subgroup of infinite group is coinfinite: For an infinite group, the complement of any proper subgroup is infinite.

Proof
Given: A group $$G$$, an automorphism $$\sigma$$ of $$G$$ that is a finitary permutation.

To prove: $$\sigma$$ is the identity automorphism.

Proof: By fact (1), the set $$C_G(\sigma)$$ of fixed points under $$\sigma$$ in $$G$$ is a subgroup of $$G$$. By the definition of finitary permutation, the complement $$G \setminus C_G(\sigma)$$ is a finite subset of $$G$$. But fact (2) tells us that if $$C_G(\sigma)$$ is proper, then $$G \setminus C_G(\sigma)$$ is infinite in size. Thus, we must have $$C_G(\sigma) = G$$, so $$\sigma$$ is the identity automorphism on $$G$$.