Let $A,B$ be $\mathbb{C}$-vector space. We can view them as a $\mathbb{Z}$-module. Suppose that there is a $\mathbb{Z}$-module isomorphism $\phi$ between $A$ and $B$. Then can we have a natural $\mathbb{Q}$(or even $\mathbb{R}$, $\mathbb{C}$)-vector space isomorphism between $A$ and $B$ from $\phi$?
I think at least we can have $\mathbb{Q}$-vector space isomorphism.
For example, Let $a\in A$ and denote $\frac{1}{n}a=\alpha\in A$ for some natural number $n$. Let $\phi$ be a $\mathbb{Z}$-module isomorphism from $A$ to $B$. Then, we have
$\phi(a)=\phi(n\alpha)=n\phi(\alpha)=n\phi(\frac{1}{n}a)$.
Thus, $\phi(\frac{1}{n}a)$ has to be $\frac{1}{n}\phi(a)$. Because $\mathbb{Q}$ is the field of fraction of $\mathbb{Z}$, we may have $\phi(\frac{m}{n}a)=\frac{m}{n}\phi(a)$ for all $a\in A$ and $\frac{m}{n}\in\mathbb{Q}$.
Am I right?
Now, I am struggling to extend to $\mathbb{R}$ or $\mathbb{C}$ vector space isomorphism.
Your argument that $\mathbb{Z}$-linear implies $\mathbb{Q}$-linear is correct but it's not possible to go beyond this. In fact assuming the axiom of choice $\mathbb{C}$ and $\mathbb{C}^2$, for example, are isomorphic as $\mathbb{Q}$-vector spaces (since they both have the same dimension, namely $|\mathbb{R}|$).
On the other hand, it's consistent with ZF that every homomorphism between Polish groups is automatically continuous, and in particular that every homomorphism between finite-dimensional real or complex vector spaces is automatically continuous. If this is true then any $\mathbb{Z}$-linear isomorphism between finite-dimensional real or complex vector spaces is automatically $\mathbb{R}$-linear, so in the real case we're done, and in the complex case they have the same real dimension and hence the same complex dimension, so there is some other $\mathbb{C}$-linear isomorphism between them.