Find $ Aut (\mathbb{C})$ , the group of automorphisms of field of complex numbers $\mathbb{C}$ .
My solution :
Take any $\phi\in Aut(\mathbb{C})$ . I think its enough if we know possible values of $\phi(i)$ because if we know $\phi(i)= a$ then we know $1=\phi(1)=\phi(i^4)=a^4$ . So $a=$ roots of equation $z^4=1$ in complex numbers .Lets name them as $a_1,a_2,a_3,a_4$
Now i can show that $\phi(x)=x a_1$ $\forall x\in \mathbb{Q}$ . I'm not sure what do about $\phi(x)$ when $x\in\mathbb{R}$ . If get possible values for $\phi(x)$ when $x\in\mathbb{R}$ i can find $\phi(z)$ for every $z\in \mathbb{C}$ ( if $z=a+ib $ $ a,b\in \mathbb{R}$ we have $\phi(z=\phi(a)+\phi(i)\phi(b)$ )
So i guess finding possible $\phi(x)$ for all $x\in \mathbb{R}$ will solve the problem . Alos i would like to share the link if its sheds some light on how to go about solving this problem
Is an automorphism of the field of real numbers the identity map?
It depends on what you mean by automorphism. Do you mean an automorphism group under composition over holomorphic functions on $\mathbb{C}$? Then your solution is $\phi(z) = \lambda z + \beta$ for all $\lambda,\beta \in \mathbb{C}$ $\lambda \neq 0$.
Do you mean a continuous automorphism of the field $\mathbb{C}$? Then your choices are $\phi(z) = z, \overline{z}$.
Do you mean an automorphism of the field $\mathbb{C}$ that is not continuous? Well then you need the Absolute Galois Group.
As you can see, your question can be interpreted many ways, it isn't clear what you are asking.
Since this is from an analysis workbook, the answer is probably the first.