Exercise with automorphisms of rational functions field

Question

Let $E $ be the field of rational functions over $\mathbb Q $ and consider the six automorphisms $$\alpha_1:f (x)\mapsto f (x), \;\alpha_2:f (x)\mapsto f (1/x)$$ $$\alpha_3:f (x)\mapsto f (1/(1-x)), \; \alpha_4:f (x)\mapsto f (1-x)$$ $$\alpha_5 : f (x)\mapsto f (x/(x-1)),\; \alpha_6:f (x)\mapsto f ((x-1)/x).$$ I have to prove that they form a group of authomorphisms $G $ (hence $|G|=6$), and that, named $F=$Inv $G $, $[E:F]\ge 6$. The first point is quite easy, however it is unclear to me why $[E:F]$ is not equal to $6$: a theorem states that if $E $ is a field and $G $ is a finite group group of authomorphisms of $E $, then $E/$Inv $G $ is a Galois extension (theorem 4.7 of Basic Algebra I). This should imply precisely$[E:F]=|G|$, so I must be wrong somewhere; thanks for any explanation.