I am new to abstract algebra and thses days studying isomorphisms.
I had to prove that if $G$ and $G^*$ are isomorphic with a isomorphism $\phi $ .
Then
$G=\langle a\rangle \Rightarrow G^*=\bigl\langle\phi (a)\bigr\rangle$. My book gives different answer but I doubt if this it could be solved this way . I went like
If $G=\langle a\rangle$ means every element of $G $ Is in the form $\{a^n\,|\, n\in \mathbb{Z}\}$.
And for every $a^n \in G$, a $\phi (a^n)\in G^*$ .
Hence all elements of $G^*$ are in form $\phi (a^n)$
And by property of isomorphism of elements $\phi (a^n)= \bigl(\phi (a)\bigr)^n$ .
Hence every element of $G^*$ is of the form $\bigl(\phi (a)\bigr)^n$
Or $G^*=\bigl\langle\phi (a)\bigr\rangle.$
Edit :
For all "a" in G there is a $\phi(a) $ in $G^*$ . So elements in G are power of a and elements in $G^*$ are the value returned by application of isomorphic function $\phi$ thus they are like $\phi(a^1),\phi(a^2)........$ but another property says that $\phi(a^n)=[(\phi(a))^n$] thus all elements in $G^*$ now can be written as power of $\phi(a)$ .
Your proof is basically correct. You should have added why you think that$$(\forall n\in\mathbb{Z}):\phi(a^n)=\phi(a)^n.\tag{1}$$Another thing missing is that $\phi$ is surjective and that's why you can deduce from $(1)$ that every element of $G^*$ is of the form $\phi(a^n)$, for some integer $n$.