Semidirect product of finite group by fixed-point-free automorphism implies all elements in its coset have same order

Statement
Suppose $$G$$ is a finite group and $$\sigma$$ is a fixed-point-free automorphism of $$G$$ of order $$n > 1$$. Let $$H$$ be the subgroup of $$\operatorname{Aut}(G)$$ generated by $$\sigma$$. Consider the external semidirect product $$G \rtimes H$$. Then, the following is true: all elements in the coset that corresponds to $$\sigma \in H$$ have order precisely $$n$$. Moreover, all elements in all cosets corresponding to cyclic generators of $$H$$ have order precisely $$n$$.

Related facts

 * Fixed-point-free involution on finite group is inverse map
 * Frobenius conjecture

Facts used

 * 1) uses::Commutator map with fixed-point-free automorphism is injective

Proof
Given: Finite group $$G$$, fixed-point-free automorphism $$\sigma$$ of $$G$$ of order $$n$$ generating a subgroup $$H$$ of $$\operatorname{Aut}(G)$$. $$K = G \rtimes H$$.

To prove: If $$k \in K$$ is of the form $$(g,\sigma), g \in G$$, then $$k$$ has order precisely $$n$$.

Proof: