IA-automorphism group of finite nilpotent group has precisely the same prime factors of order as the derived subgroup

Statement
Suppose $$G$$ is a finite nilpotent group. Then, the IA-automorphism group of $$G$$ (i.e., the subgroup of the automorphism group of $$G$$ comprising the IA-automorphisms -- those automorphisms that preserve each of the cosets of the derived subgroup) is a finite group and the prime factors of its order are precisely the same as the prime factors of the order of the derived subgroup of $$G$$.

Facts used

 * 1) uses::Equivalence of definitions of finite nilpotent group
 * 2) uses::IA-automorphism group of finite p-group is p-group

Proof
Given: A finite nilpotent group $$G$$.

To prove: The prime factors of the order of the IA-automorphism group of $$G$$ are precisely the same as the prime factors of the order of the derived subgroup of $$G$$.

Proof: By Fact (1) and some more work, we can see that an automorphism of $$G$$ is an IA-automorphism if and only if it induces IA-automorphisms on each of the Sylow subgroups of $$G$$. We can work out from this that the IA-automorphism group of $$G$$ is isomorphic to the external direct product of the IA-automorphism groups of each of the Sylow subgroups of $$G$$. By Fact (2), for each prime factor $$p$$ of $$G$$, the corresponding IA-automorphism group is a $$p$$-group, so the IA-automorphism group of $$G$$ is isomorphic to a direct product of groups whose orders are powers of primes appearing in the prime factorization of $$G$$.

It remains to show that a prime appears with a positive power if and only if it also appears as a factor in the order of the derived subgroup. This follows from the observation that, for a prime number $$p$$, the IA-automorphism group for the $$p$$-Sylow subgroup is trivial if and only if the corresponding $$p$$-Sylow subgroup in $$G$$ is abelian, which in turn is equivalent to saying that the derived subgroup of $$G$$ does not have $$p$$ dividing its order. This completes the proof.