a)Describe the automorphism group of $\operatorname{Aut}(\mathbb{Z}/9\mathbb{Z})$.
b) Prove that if a Group G has the trivial center then $|\operatorname{Aut}(G)| \geq |G|$.
My attempt: a) Clearly, $\mathbb{Z}/9\mathbb{Z}$ is a cyclic group.
We know that the automorphism group of the cyclic group of order n is isomorphic to $(\mathbb{Z}/9\mathbb{Z})^{*}$, and an abelian group of order $\phi(n)$.
This implies $\operatorname{Aut}(\mathbb{Z}/9\mathbb{Z})$ is isomorphic to $(\mathbb{Z}/9\mathbb{Z})^{*}$.
Let $G=(\mathbb{Z}/9\mathbb{Z})^{*}$ and $|G|=\phi(9)=6$. By classification of order of 6 group, we have $G$ is isomorphic to $\mathbb{Z}/6\mathbb{Z}$ because $G$ is abelian and hence it is cyclic.
b)We know that $\operatorname{Inn}(G) \simeq G/Z(G)$, where $Z(G)$ denote the center of G.
Since $Z(G)=\{e\}$, then $\operatorname{Inn}(G) \simeq G$, and we know that $\operatorname{Inn}(G)$ is a normal subgroup of $\operatorname{Aut}(G)$.
This implies $|\operatorname{Inn}(G)| \leq |\operatorname{Aut}(G)|$, i.e, $|G| \leq |\operatorname{Aut}(G)|$.
Is both part of the question correct? This is first time for me to study this topic. Please suggest me any improvement in this solution. Many thanks in advance.
Both answers are correct. Well done!
The intuition behind $(b)$ is that your group is maximally unabelian so conjugation gives for every choice of element a new inner automorphism.