If $p$ is an odd prime then prove that there cannot exist a finite group $G$ such that ${\rm Aut}(G)\cong \mathbb{Z}_p$. 
If $p$ is an odd prime then prove that there cannot exist a finite group $G$ such that ${\rm Aut}(G)\cong \mathbb{Z}_p$.

Can anyone tell me how to proceed in this question?
Here is my attempt:
If ${\rm Aut}(G)$ is cyclic then ${\rm Inn}(G)$ is also cyclic which implies $G/Z(G)$ is cyclic which implies $G$ is abelian.
If $G$ is abelian then for $\phi \in{\rm Aut}(G)$ such that $\phi(g)=g^{-1}$. then $\phi(\phi(g)) = \phi(g^{-1}) = g$. Hence $\phi$ is of order $2.$
Hence the order of ${\rm Aut}(G)$ must be a multiple of 2 and hence cannot have prime order.
Edit:- I cannot seem to proceed when $\phi$ becomes just the identity mapping. In that case every element of the group has order $2$ (except for the identity element of order $1$) .

Hint: yes $G$ is abelian and consider the map $\phi$: $G \rightarrow G$, defined by $\phi(g)=g^{-1}$. Prove that this is an isomorphism. What is the order of $\phi$?
