For any group $G$, you have a bijection between $G$ and the set $Hom(\mathbb{Z},G)$ of group homomorphisms $\mathbb{Z}\to G$.
The bijection is as follows: for $x\in G$, set $h_x:\mathbb{Z}\to $G$, \ m\mapsto x^m$.
Then the desired bijection is $G\to Hom(\mathbb{Z},G), \ x\mapsto h_x.$
First of all, you have to check that $h_x$ is indeed a group homomorphism (easy).
For the injectivity part: if $x,y\in G$ are such that $h_x=h_y$, then $h_x(1)=h_y(1)$, that is $x=y$?
For the surjectivity part: the main idea is that any element of $\mathbb{Z}$ is the sum of several copies of $1$ (or $-1$). Thus, in order to define a homomorphism $\varphi: \mathbb{Z}\to G$, it is enough to know $\varphi(1)$.
More precisely, if $\varphi$ is such a morphism, then, for all $m\geq 0$, we have $\varphi(m)=\varphi(1+\cdots+1)=\varphi(1)^m$.
If $m<0$, then $\varphi(m)=\varphi(-(-m))=\varphi(-m)^{-1}$ (a morphism repects inverses), so $\varphi(m)=(\varphi(1)^{-m})^{-1}=\varphi(1)^m$.
Finally, $\varphi=h_x$ with $x=\varphi(1)$.
In particular, if $G=\mathbb{Z}_n$, you get $n$ such morphisms.
Concerning your second question, you have only the trivial group/ring morphism. Any element of $\mathbb{Z}_n$ has finite order. But a morphism sends an element of finite order to an element of finite order. Since the only element of $\mathbb{Q}$ of finite order is $0$....
Edit Some answers were given while I was typing. Feel free to delete this post if it seems useless.
Best Answer
Yes, your description is correct, and moreover it uses no special features of $\mathbb{C}$: this is a description of homomorphisms from $\mathbb{Q}(x)$ to any field, which exhibits the universal property of $\mathbb{Q}(x)$ in the category of fields, namely, it is the free field on a transcendental element.