Let $G$ be a compact monothetic group, that is, $G$ is a compact group with a dense cyclic subgroup. Note that $G$ is necessarily abelian. Suppose that $$G=\overline{\{a^{k}:k\in\mathbb{Z}\}}=\overline{\{b^{k}:k\in\mathbb{Z}\}}.$$ Can we find a topological group isomorphism (i.e. isomorphism + homeomorphism) $\phi\colon G\to G$ such that $\phi(a)=b$?
Without the closures and topology, I think one can define $\phi(a^{k}):=b^{k}$. However, I'm not sure if this injective and that this can be extended continuously to the closure.
No. For instance, let $G=\mathbb{R}/\mathbb{Z}$. Then any irrational number generates a dense cyclic subgroup, but the only topological automorphisms of $G$ are the identity and the negation map.
More generally, Pontryagin duality is very useful for understanding questions like this. A homomorphism $\mathbb{Z}\to G$ with dense image is dual to an injective homomorphism $\hat{G}\to\mathbb{R}/\mathbb{Z}$ where $\hat{G}$ is the Pontryagin dual. So your question is, if $H=\hat{G}$ is a subgroup of $\mathbb{R}/\mathbb{Z}$, then are any two injective homomorphisms $H\to\mathbb{R}/\mathbb{Z}$ related by an automorphism of $H$? The answer is obviously no since you could have injective homomorphisms $H\to\mathbb{R}/\mathbb{Z}$ that do not even have the same image (as in the example above, where $H=\mathbb{Z}$ and any irrational number gives an injective homomorphism $\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$).