Commutator map with fixed-point-free automorphism is injective

Statement
Suppose $$G$$ is a group and $$\sigma$$ is a fixed-point-free automorphism of $$G$$, i.e., the only fixed point of $$\sigma$$ is the identity element. Then, the mapping:

$$x \mapsto x\sigma(x^{-1})$$

is an injective self-map from $$G$$ to itself.

Applications

 * Fixed-point-free involution on finite group is inverse map
 * Semidirect product of finite group by fixed-point-free automorphism implies all elements in its coset have same order

Proof
Given: Group $$G$$, fixed-point-free automorphism $$\sigma$$, elements $$x,y \in G$$ such that $$x\sigma(x^{-1}) = y\sigma(y^{-1})$$

To prove: $$x = y$$

Proof: With some manipulation, we can rewrite the condition as:

$$y^{-1}x = \sigma(y^{-1})(\sigma(x^{-1})^{-1}$$

Simplify the right side using that $$\sigma$$ is an automorphism, and obtain:

$$y^{-1}x = \sigma(y^{-1}x)$$

Since $$\sigma$$ is fixed-point-free, and $$y^{-1}x$$ is a fixed point of $$\sigma$$, this forces that $$y^{-1}x$$ is the identity element of $$G$$, yielding $$x = y$$.