I solved at
For $f$ in ${\rm Gal}(\Bbb C/\Bbb Q),$
$F(i)$ is $i$ or $-i$
And for $a,b$ in $\Bbb Q$, $f(a+bi)=a+bf(i)$.
I have to classify $G(\Bbb C/\Bbb Q)$, but I find the automorphism $f$ is defined on $\Bbb Q[i]$, that is the coefficient $a,b$ of $a+bi$ is rational...
How can I extended for irrational?
Given what you said in the comment, what you actually want to do is show that $\mathbb{C}/\mathbb{Q}$ is Galois in the sense that $\mathbb{C}^{\operatorname{Aut}(\mathbb{C}/\mathbb{Q})} = \mathbb{Q}$. In other words, you want to show that if $z\in \mathbb{C}\setminus \mathbb{Q}$, there is an automorphism $\varphi$ of $\mathbb{C}$ such that $\varphi(z)\neq z$. This most definitely does not require to classify all automorphisms of $\mathbb{C}$.
As a first comment, this relies heavily on the axiom of choice: you can find models of ZF where the only non-trivial automorphism of $\mathbb{C}$ is the conjugation. This being said, once we don't mind using transcendence bases, we can answer your question.
If $z$ is algebraic over $\mathbb{Q}$, write $\mathbb{Q}\subset K\subset \mathbb{C}$ such that $K/\mathbb{Q}$ is purely transcendental and $\mathbb{C}/K$ is algebraic. Then $\mathbb{C}$ is the algebraic closure of $K$ and $z$ is algebraic over $K$, so there is an automorphism $\phi\in \operatorname{Gal}(\mathbb{C}/K)$ such that $\varphi(z)\neq z$.
If $z$ is transcendental over $\mathbb{Q}$, then we can choose a transcendence basis $S$ of $\mathbb{C}/\mathbb{Q}$ with $z\in S$. Then there is an automorphism of $\mathbb{Q}(S)$ swapping $z$ with another element of $S$. This automorphism can be extended to $\mathbb{C}$ since it is the algebraic closure of $\mathbb{Q}(S)$.