Base change through a field automorphism 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...)
 A: let me share my thoughts on your question. For clarity, let us work first with affine $n$-space; a point is then just a Morphism $p: \mathbb{C}[T_1, \ldots, T_n] \to \mathbb{C}$ of $\mathbb{C}$-algebras with kernel $(T_1 - p_1, \ldots, T_n-p_n)$. 
The base change map $X^\sigma \to X$ is then $\mathbb{C}[T_1, \ldots, T_n] \to \mathbb{C}[T_1, \ldots, T_n] \otimes_{\sigma^{-1}} \mathbb{C}, P \mapsto P \otimes 1$ - note, the important thing is the different $\mathbb{C}$-algebra-structure on the right, where $\lambda \in \mathbb{C}$ is sent to $1 \otimes \lambda$. With this in mind, the map $$
\mathbb{C}[T_1, \ldots, T_n] \otimes_{\sigma^{-1}} \mathbb{C} \to \mathbb{C}[T_1, \ldots, T_n], P\otimes \lambda \to \lambda (P)^{\sigma^{-1}},
$$ 
where $\sigma^{-1}$ acts on the coefficients, is an isomorphism of $\mathbb{C}$-algebras.
Rephrasing again: On the level of rings, $X^\sigma \to X$ is given by "$P \mapsto P^{\sigma^{-1}}$" for polynomials, if we identify both $X$ and $X^\sigma $ with $Spec(\mathbb{C}[T_1, \ldots, T_n])$. Now consider a point of $X^\sigma $, that ist, a map $p$ as above, its image under the geometric base change map corresponds to the pullback on the level of rings, that is, its image in $X$ is the composite map $\mathbb{C}[T_1, \ldots, T_n] \to \mathbb{C}, P \mapsto p(P^{\sigma^{-1}})$ and we see that its kernel is given by $(T_1 - \sigma(p_1), \ldots, T_n - \sigma(p_n))$
Now i only considered affine n-space instead of projective varieties, but this is really the key point - closed subschemes of it work analogously, the key algebra isomorphism just becomes
$$
\mathbb{C}[T_1, \ldots, T_n]/(f_1, \ldots, f_m) \otimes_{\sigma^{-1}} \mathbb{C} \to \mathbb{C}[T_1, \ldots, T_n]/(f_1^{\sigma^{-1}}, \ldots, f_m^{\sigma^{-1}})
$$ 
Glueing the morphisms should now give you the desired projective identity.
I apologize if the answer is not right on point, but i wanted to share the affine considerations, i hope they help you.
Best regards.
