Determine all the field automorphism of $\mathbb{Q}(x)$ of order two.
I know that every field automorphism must be the identity map over $\mathbb{Q}$, and so they are uniquely determined given the image of $x$. For example consider $$ \varphi: \mathbb{Q}(x) \to \mathbb{Q}(x) \quad \varphi(x) = \frac{1}{x} $$ Then $\varphi$ is a well defined automorpfism and $\varphi \circ \varphi (x) = x$ which means it has order two.
I've alredy proved that in general if $\varphi(x) = \frac{P(x)}{Q(x)}$, then $$\deg P - \deg Q = \pm 1$$
And also that $x \to -x$ and $x \to -\frac{1}{x}$ are two trivial automorphisms. Can you help me saying something more?
Each automorphism of $\Bbb Q(x)$ satisfies $$\phi(x)=\frac{ax+b}{cx+d}$$ with $ad-bc\ne0$. Associating this with the matrix $A=\pmatrix{a&b\\c&d}$ shows the automorphism group is isomorphic to $\text{PSL}_2(\Bbb Q)$. This represents a degree two automorphism iff $A$ is not a scalar matrix but $A^2$ is. This boils down to $A$ having trace zero, that is $d=-a$.