Towards the solution of the Problem : Field Extension problem beyond $\mathbb C$ (Question 1) I am posting this problem in order to break the problem in my previous post  Field Extension problem beyond $\mathbb C$.
Notation: $M(\mathbb C):=$ Field of all meromorphic functions on $\mathbb C$,
 $K:=$ a proper subfield between $\mathbb C$ and $M(\mathbb C)$
Question 1: Let $\sigma\in Aut(K)$ and $c$ is a constant function. Is it true $\sigma(c)$ is again a constant function ?
I will post subsequent questions after getting a good answer (either by myself or someone else). Let us try how far we can go! 
 A: Yes, it is true that for $c\in\mathbb{C}$ then $\sigma(c)$ is again in $\mathbb{C}$. This is because the subfield $\mathbb{C}$ of $K$ is distinguished by the following algebraic property.

An element $c\in K$ is in $\mathbb{C}$ if and only if $X^n-c+q=0$ has a root in $K$ for each $n\in\mathbb{N}$ and $q\in\mathbb{Q}$.

Being an automorphism of $K$, $\sigma$ preserves the stated property, so maps elements of $\mathbb{C}$ (or, 'constant functions') into $\mathbb{C}$.
The fact that each $c\in\mathbb{C}$ satisfies the stated property above follows from the fact that $\mathbb{C}$ is algebraically closed. On the other hand, if $c\in K\setminus\mathbb{C}$ is a nonconstant meromorphic function then, by the little Picard theorem, it maps onto $\mathbb{C}$ minus at most two points, so its image will contain (infinitely many) rationals $q$ (or you can use the simpler fact that the image of a nonconstant meromorphic function is a connected dense open subset of $\mathbb{C}$). So, $c-q$ has a zero,  which must be of a finite order $m\ge1$. Then, $c-q$ does not have an $n$'th root (in the field of meromorphic functions) for any $n$ not dividing $m$.
