Commutator of finite group with cyclic coprime automorphism group equals second commutator

Statement
Let $$G$$ be a finite group and $$A$$ be a cyclic subgroup of $$\operatorname{Aut}(G)$$ such that the orders of $$G$$ and $$A$$ are relatively prime. Then:

$$[[G,A],A] = [G,A]$$.

Facts used

 * 1) uses::Centralizer-commutator product decomposition for finite groups
 * 2) uses::Cyclicity is quotient-closed
 * 3) uses::Lagrange's theorem
 * 4) uses::Order of quotient group divides order of group