My question comes from the reading of the first two pages of the article "Bernhard Köck - Belyi's Theorem Revisited".
Consider a field $C$ and a variety $X$ over $C$. In particular $X$ is a separated, integral scheme with a structure morphism $f:X\longrightarrow \operatorname{Spec}C=\{(0)\}$ of finite type (in this case the morphism $f$ is the constant morphism that sends $X$ to $(0)$). If $\sigma\in\operatorname{Aut}C$, the author defines the variety $X^\sigma$ over $C$ in the following way:
As scheme, the variety $X^\sigma$ is $X$, but the structure morphism is $\operatorname{Spec} (\sigma)\circ f: X\longrightarrow\operatorname{Spec} C$
Now since $\operatorname{Spec} C$ is a point, it is evident that $\operatorname{Spec}(\sigma):C\longrightarrow C$ is the identity wich sends $(0)$ to $(0)$ and it follows that $\operatorname{Spec} (\sigma)\circ f=f$. For this reason for me $X$ and $X^\sigma$ are the same object! I'dont understand the difference between the varieties $X$ and $X^\sigma$.
A morphism of schemes is not determined by the underlying map on topological spaces! There is also a map of sheaves to consider.
For example, take $C = X=\mathbb{C}$, and let $\sigma: \mathbb{C}\to\mathbb{C}$ be complex conjugation. Then $\sigma$ gives rise to a morphism of schemes $\operatorname{Spec}\mathbb{C}\to\operatorname{Spec}\mathbb{C}$ which is different from the identity map—even though they both are the same map of one-point topological spaces.