Vector Space isomorphisms of $\mathbb{Q}(z)$ preserving the Galois group (where $z$ is a primitive third root of unity) Take the field extension $\mathbb{Q}(z)$ where $z$ is a primitive third root of unity and consider the set $A$ of vector-space automorphisms of $\mathbb{Q}(z)$ so that for $T \in A$ the map $\phi \mapsto T\phi T^{-1}$ is an isomorphism of the Galois group. It is easy to show that $T \phi T^{-1}$ is an automorphism of the additive group of $\mathbb{Q}(z)$ since $T$ and $\phi$ are both linear. However, showing that $T \phi T^{-1}(ab)=T \phi T^{-1}(a) T \phi T^{-1}(b)$ for each automorphism $\phi$ and $a, b \in \mathbb{Q}(z)$ puts a condition on $T$.
Specifically, I can show using other methods that any $T$ fixes the degree of the minimal polynomial of elements of $\mathbb{Q}$ and so $T$ must be of the form $1 \mapsto c, z \mapsto a+bz$ where $b \neq 0$ and $a, b, c \in \mathbb{Q}$. Then the multiplicative condition on $T \phi T^{-1}$ gives the condition $a+c+ac=b$. Thus, it appears that maps of the form $1 \mapsto c, z \mapsto a+(a+c+ac)z$ should work. Unfortunately, my computations show that such maps do not form a group and yet, from my definition of these maps above, it seems like they should. What am I doing wrong here? 
 A: Complex conjugation $\phi:z\mapsto z^2=-1-z$ is the only non-trivial automorphism of $\mathbb{Q}(z)$, so we must have $T\phi T^{-1}=\phi$, i.e. $T$ must commute with $\phi$.
As a linear transformation of vector space over the rationals $\phi$ is semisimple, and it has eigenvalues $+1$ (resp. $-1$) with the real (resp. imaginary line) being the eigenspaces. A basic fact of commuting linear transformations is that they must preserve the eigenspaces. In other words, we must have $T(1)=c\in \mathbb{Q}, c\neq0$ (as observed by the OP, also) as well as $T(\sqrt{-3})=a\sqrt{-3}$ for some $a\in\mathbb{Q}, a\neq0$. 
It turns out that these conditions are also sufficient. This is perhaps easiest to see by writing the linear matrices of these linear transformations w.r.t. the eigenbasis $\{1,\sqrt{-3}\}$. We have
$$
M(\phi)=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right),\quad
\text{and}\quad
M(T)=\left(\begin{array}{cc}c&0\\0&a\end{array}\right).$$
It is clear that these matrices commute irrespective of the choice of $a,c\in\mathbb{Q}^*$. Earlier we showed that these are the only possibilities, so the question is solved in this case.
But do observe that this thinking doesn't immediately generalize to cubic Galois extensions. One difference comes from the possibility that conjugation by $T$ does not need to be the trivial automorphism of the Galois group. Another difference comes from the fact that the smaller field no longer contains the eigenvalues of the matrices of the elements of the Galois group. A different approach will be needed.
