Let $L=\mathbb{Q}(x)$ be the field of rational functions over $\mathbb{Q}$ and let $\sigma,\tau\in\text{Aut}(L)$ be given by $\sigma(f(x))=f(1/x)$ and $\tau(f(x))=f(1-x)$ for $f(x)\in\mathbb{Q}(x)$. Determine the field of invariants for both of these automorphisms. That is, determine $L^{\langle\sigma\rangle}$ and $L^{\langle\tau\rangle}$.
I'm a bit new to field and Galois theory but this is my attempt. We have that $L^{\langle\sigma\rangle}=\{f(x)\in\mathbb{Q}(x):\alpha(f(x))=f(x), \forall\alpha\in\langle\sigma\rangle\}$. So then I calculated what all the elements of ${\langle\sigma\rangle}$ are, but $\sigma^2(f(x))=\sigma(f(1/x))=f(x)$. So I think that $\langle\sigma\rangle$ has only two elements, namely $\sigma$ and $\sigma^2=id$. The same goes for elements of ${\langle\tau\rangle}$ because $\tau^2(f(x))=\tau(f(1-x))=f(x)$. I solved my first confusion (thanks to very lazy notation) but now I'm a bit confused. Aren't both fixed fields just the same one since they have only two elements respectively?
It is a classical result in Galois theory that $L/L^{\langle\sigma\rangle}$ is an extension (actually, even a Galois extension) of degree $|\langle\sigma\rangle|=2$. Note that $x+\frac{1}{x}$ clearly belongs to the fixed field, and so $\mathbb{Q}(x+\frac{1}{x})\subseteq L^{\langle\sigma\rangle}$. Also note that $x$ is a root of the polynomial $T^2-T(x+\frac{1}{x})+1\in\mathbb{Q}(x+\frac{1}{x})[T]$, and hence:
$[L:\mathbb{Q}(x+\frac{1}{x})]\leq 2$
It follows that $\mathbb{Q}(x+\frac{1}{x})$ can't be strictly contained in $L^{\langle\sigma\rangle}$, as otherwise the extension degree would be larger than $2$. Hence $L^{\langle\sigma\rangle}=\mathbb{Q}(x+\frac{1}{x})$.
Now do something similar to find $L^{\langle\tau\rangle}$.