Isomorphism of $(\Bbb{Z}, *)$ and $(\Bbb{Q}-\{0\}, \cdot)$ Are there any operations on $\Bbb{Z}$ that makes it isomorphic to $(\Bbb{Q}-\{0\}, \cdot)$ as a group? 
Edit: the operation should be made of addition and multiplication of integers, possibly recursively. If there is no such iso, then why?
 A: Of course! You could just take some bijection between $\mathbb{Q}\setminus\{0\}$ and $\mathbb{Z}$ and artificially make it a homomorphism.
You then get the group structure for free, because you made sure it was a homomorphism and because it is a bijection. Bijections are just re-labellings, so that is all that is going on. If I relabel $3/5$ as "$7$", $2/3$ as "$12$" and $2/5$ as "$1729$" then $7\ast 12=1729$.
A: Cardinality is the only invariant of the underlying set of a group that matters. 
If $(G, \circ)$ is a group and you have a bijection of sets $f: G \rightarrow T$, then you can define a group operation $*$ on $T$ so that $f$ is an isomorphism. In fact the operation $*$ is determined by the requirement that $f$ should be an isomorphism.
If $f$ is an isomorphism, then $f(x \circ y) = f(x) * f(y)$. Then 
$$x * y = f(f^{-1}(x)) * f(f^{-1}(y)) = f(f^{-1}(x) \circ f^{-1}(y))$$
Of course, you should still check that for any bijection $f$ the operation $*$ above makes $(T, *)$ into a group. 
This might seem a bit artificial, but you can see that all group structures $(T, *)$ that are isomorphic to $(G, \circ)$ arise in this manner.
I can see after the edit that this does not really answer your intended question: whether there is a "natural way" (eg using multiplication and addition of integers) to make $(\mathbb{Z}, *)$ isomorphic to the multiplicative group of nonzero rationals.
