Ortbit of $\pi$ under the action of $\operatorname{Aut} \mathbb C $ It is well known that if we assume Zorn lemma, then $\operatorname{Aut} \mathbb C=\operatorname{Gal}(\mathbb C/\mathbb Q)$ is a very big group. Moreover it is very difficult to express explicitly any element of $\operatorname{Aut} \mathbb C$ different from conjugation.


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*What is the set $\{\sigma(\pi)\colon\sigma\in\operatorname{Aut} \mathbb C \}$? Is it isomorphic to $\mathbb C$?

*Is it possible to choose $z\in\mathbb C$ such that $\{\sigma(z)\colon\sigma\in\operatorname{Aut} \mathbb C \}=\mathbb C$?

 A: Here's an approach to the question :
$\mathbb{C}$ is algebraic over $\mathbb{Q}(S)$ where $S$ is a transcendance basis. If you recall the proof of the existence of a transcendance basis, then you notice that it uses Zorn's lemma and that we can actually make it so that $\pi\in S$. In particular, any permutation of $S$ induces an automorphism of $\mathbb C$ ($\mathbb C$ is algebraically closed). Therefore for any element $s$ of $S$ you have an automorphism of $\mathbb C$ that sends $\pi\mapsto s$. Thus the orbit of $\pi$ contains $S$. 
A first corollary is that the orbit of $\pi$ has the cardinality of the continuum (as does $S$). So it is "isomorphic to $\mathbb C$ " if you're referring to size.
Moreover let $z\in \mathbb{C}$ be a complex number that is algebraically independent of $\pi$. Then by the same remark as above, there is $S$ containing both $\pi$ and $z$, and so there is an automorphism sending $\pi\to z$.  Thus the orbit contains $\pi$ and every algebraically independent complex number. 
Of course thus argument works when replacing $\pi$ by any transcendental complex number; and doesn't for algebraic complex numbers (whose orbit has size the degree of their minimal polynomial over $\mathbb{Q}$).
But we can go further : suppose you have two transcendental numbers, $z,w$. Then there are transcendence bases $S_1, S_2$ such that $z\in S_1, w\in S_2$. Moreover, these two transcendence bases have the same cardinality, so there is a bijection $f: S_1\to S_2$ sending $z\mapsto w$, which induces an isomorphism $\mathbb{Q}(S_1)\to \mathbb{Q}(S_2)$, which extends to $\mathbb{C\to C}$ ($\mathbb{C}$ is algebraically closed) and thus sends $z\mapsto w$. Thus the orbit of any transcendental complex number is the set of all transcendental complex numbers.
Therefore we have three types of orbits :
1- the rationals, they have a singleton orbit, they are fixed by any automorphism.
2- the algebraic complex numbers, their orbit is the set of all roots of their minimal polynomial over $\mathbb{Q}$
3- The transcendental complex numbers, which form one orbit altogether.
Now your question 2. has an obvious answer which is no: if $\sigma (z) =1$, then $z=1$. The most you can ask for is if there's $z$ such that its orbit is precisely the set of transcendental complex numbers, which is what we proved earlier.
