Suppose that we have Galois covering $$X\longrightarrow P^{1}_{\mathbb{Q}}$$ defined over the rationals which is cyclic, in the sense that the Galois group associated to the covering is a cyclic group. Then $X$ can not be $P^{1}_\mathbb{Q}$. I read this result somewhere. Is this result true? I tried this over the complex numbers and I found that over complex numbers it is true:
$P^{1}_{\mathbb{C}} \longrightarrow P^{1}_{\mathbb{C}}$ defined by $z$ going to $z^{n}$.
Your (!) MO-question (together with the answer/comments) shows that the group of order $3$ is also a Galois group of a Galois cover $\mathbb P^1_{\mathbb Q}\to\mathbb P^1_{\mathbb Q}$. Besides this and order $2$, the only other possible order is $4$ which can be seen as follows:
As $\text{Aut}_{\mathbb Q}(P^1_{\mathbb Q})=\text{PGL}_2(\mathbb Q)$, your question reduces to finding elements of finite order in $\text{PGL}_2(\mathbb Q)$. Let $A\in\text{GL}_2(\mathbb Q)$ represent an element of order $m$ in $\text{PGL}_2(\mathbb Q)$. Then $A$ is diagonalizable over a quadratic extension of $\mathbb Q$. Let $a$ and $b$ be the eigenvalues of $A$. Then $\zeta=a/b$ is a primitive $m$-th root of unity. Together with the fact that $\zeta$ has degree at most $2$ over the rationals, one gets $m\le4$ (e.g. by using the irreducibility of the cyclotomic polynomials, or by more elementary means).
The order $4$ indeed arises, as $z\mapsto(z-1)/(z+1)$ has order $4$ on $\mathbb P^1_{\mathbb Q}$.
