**Note**:For a correct comprehension of the question see the "important edit" at the end.

Consider a projective variety over $\mathbb C$, $X=\textrm{Proj}\frac{\mathbb C[T_0,\ldots,T_n]}{f_1,\ldots,f_m}$ and a field automorphism $\sigma\in Aut(\mathbb C)$. Now with $X^\sigma$ we indicate the base change of $X$ along $\textrm{Spec}\,\left(\sigma^{-1}\right):\textrm{Spec}\mathbb C\longrightarrow \textrm{Spec}\mathbb C$. $X^\sigma$ as scheme is equal to $X$, but the structural morphism that gives the structure of $\mathbb C$-scheme is $(\textrm{Spec}\sigma)\circ p$.

I have proved that $X^\sigma\cong\textrm{Proj}\frac{\mathbb C[T_0,\ldots,T_n]}{f^{\sigma^{-1}}_1,\ldots,f^{\sigma^{-1}}_m}$ ($f^{\sigma^{-1}}_1$ means that change the polynomial coefficients of $f_i$ by applying $\sigma^{-1}$). Therefore if we think in terms of projective algebraic sets: $X$ is $Z(f_1,\ldots,f_m)\subseteq\mathbb P^n(\mathbb C)$ and $X^\sigma $ is $Z(f^{\sigma^{-1}}_i,\ldots,f^{\sigma^{-1}}_m)\subseteq\mathbb P^n(\mathbb C)$.

Now I don't understand why the unique scheme isomorphism $X^\sigma\longrightarrow X$ induced by the fidebered product corresponds to the following map:

$$Z(f^{\sigma^{-1}}_1,\ldots,f^{\sigma^{-1}}_m)\longrightarrow Z(f_1,\ldots,f_m)\quad\quad(\ast)$$

$$P\longmapsto\sigma(P)$$

At level of schemes this map is simply the identity of the scheme $X$ so I don't understand the role of $\sigma$.

**Important Edit:** As Felipe Voloch says in the comments, I have merged two different definitions of "twist" and in the above diagram, the presence of the map "$\text{id}$" **is wrong**. I'd like more informations about the difference of the following two ways to obtain of $X^\sigma$:

- One $X^\sigma$ is obtained by changing only the structural morphism of $X$ (see the article B.Koeck - Belyi Theorem revisited at notation 1.1 for this construction).
- Another $X^\sigma$ comes from the base change induced by $\text{Spec}(\sigma)$ (
**I'm interested to this construction**).

However my original question remains, I reformulate it for clarity: consider $X^\sigma=X\times_{\text{Spec}(\mathbb C)}\text{Spec}(\mathbb C)$, I need a formal proof of the fact that the canonical map $X^\sigma\longrightarrow X$ corresponds to $(\ast)$ through the functor that relates algebraic sets and $\mathbb C$-schemes (reduced, separated...)