Suppose that $\phi$ is a topological isomorphism between the topological groups $G=\overline{\{x^{n}:n\in\mathbb{Z}\}}$ and $H=\overline{\{y^{n}:n\in\mathbb{Z}\}}$. Can we conclude that $\phi(x)=y$?
Since $\phi$ is a homeomorphism, I think that we can deduce that $\phi(x)$ also generates $H$: $$H=\phi(G)=\phi(\overline{\{x^{n}:n\in\mathbb{Z}\}})=\overline{\phi(\{x^{n}:n\in\mathbb{Z}\})}=\overline{\{\phi(x)^{n}:n\in\mathbb{Z}\}}.$$ But is this sufficient to conclude that $\phi(x)=y$?
The answer is no, because generators need not be unique. Give the group $\mathbb{Z}$ the discrete topology. Then $\phi: \mathbb{Z} \to \mathbb{Z}$ by $\phi(x)=-x$ is an isomorphism in the category of topological groups. However, if the domain and codomain were taken to be $\langle 1 \rangle$ then $\phi$ does not map fixed generator to fixed generator.
Note that $\phi(1)=-1$ does in fact generate the codomain $\mathbb{Z}$, but this is not the "chosen" generator at the outset. The problem is, of course, that generators are not unique (even in the monothetic situation).