Field automorphisms and varieties Let $C$ be an algebraically closed field and consider a variety $X$ over $C$. In the language of schemes $X$ is a separated, integral scheme over $C$ with a morphism of finite type
$f:X\longrightarrow \operatorname{Spec} C$. If $\sigma\in\operatorname{Aut} C$ then we can define the variety $X^\sigma$ as the scheme $X$ but with a morphism of finite type $\operatorname{Spec}(\sigma)\circ f: X\longrightarrow \operatorname{Spec} C$. So $X$ and $X^\sigma$ are the same as schemes but not as varieties.
Now I'm trying to investigate the relationship between $X$ and $X^\sigma$ aside from the scheme theory, but in the language of classical algebraic geometry. In literature I've found the following theorem with a very short proof that I  don't understand.

Theorem: Let $V(f_1,\ldots,f_m)\subset\mathbb P^n_C$ a classical variety where $f_1,\ldots,f_m$ are homogeneous polynomial of $C[x_0,\ldots,x_n]$. If $\sigma\in\operatorname{Aut(C)}$, and with an abuse of notation  we also indicate with $\sigma$ the induced authomorphism on $C[x_0,\ldots,x_n]$, then the closed subset $V(\sigma^{-1}(f_1),\ldots,\sigma^{-1}(f_m))$ "corresponds" to $X^\sigma$ (defined above) in the category of classical varieties. To be more precise if we apply the usual functor to the category of schemes at $V(f_1,\ldots,f_m)$ we obtain a variety $X$ and if we apply the same functor at $V(\sigma^{-1}(f_1),\ldots,\sigma^{-1}(f_m))$ we obtain $X^\sigma$.

Proof: The two graded algebras $\frac{C[x_0,\ldots,x_n]}{(f_1,\ldots,f_m)}$ and $\frac{C[x_0,\ldots,x_n]}{(\sigma^{-1}(f_1),\ldots,\sigma^{-1}(f_m))}$ are clearly isomorphic thanks to an isomorphism $\widetilde\sigma$ induced by $\sigma$. Using  the morphism $\operatorname{Proj}(\widetilde\sigma)$ follows easily the thesis.

Practically the theorem explains what is $X^\sigma$ for concrete varieties such as  closed subsets of $\mathbb P^n_C$, but I don't understand the last sentence of the proof; in particular applying the functor $\operatorname{Proj}$, I see only the isomorphism between $\operatorname{Proj}\left(\frac{C[x_0,\ldots,x_n]}{(f_1,\ldots,f_m)}\right)$ and $\operatorname{Proj}\left(\frac{C[x_0,\ldots,x_n]}{(\sigma^{-1}(f_1),\ldots,\sigma^{-1}(f_m))}\right)$.
Maybe it is a stupid question, since the proof should follow easily, but please help me to understand it. 
Thanks in advance.

References: Essentially two articles: "B.Köck - Belyi's theorem revisited" and "H.Hammer, F.Herrlich - a remark on the moduli field of a curve "
 A: This is not a complete answer, but I will describe what happens in the affine case. Suppose we have polynomials $f_1,\ldots,f_m\in C[y_1,\ldots,y_n]$ and an automorphism $\sigma: C\rightarrow C$. Then we have a commutative diagram
$$\begin{array}[c]{ccc}
C&\rightarrow&\frac{C[y_1,\ldots,y_n]}{(f_1,\ldots,f_m)}\\
\uparrow\scriptstyle{\sigma}& &\uparrow\scriptstyle{\sigma}\\
C&\rightarrow& \frac{C[y_1,\ldots,y_n]}{(\sigma^{-1}f_1,\ldots,\sigma^{-1}f_m)}
\end{array}
$$
which induces a conmutative diagram
$$\begin{array}[c]{ccc}
Spec(C)&\leftarrow&Spec\left(\frac{C[y_1,\ldots,y_n]}{(f_1,\ldots,f_m)}\right)\\
\downarrow\scriptstyle{\sigma}& &\downarrow\scriptstyle{\sigma}\\
Spec(C)&\leftarrow& Spec\left(\frac{C[y_1,\ldots,y_n]}{(\sigma^{-1}f_1,\ldots,\sigma^{-1}f_m)}\right).
\end{array}
$$
Then $\sigma:Spec\left(\frac{C[y_1,\ldots,y_n]}{(f_1,\ldots,f_m)}\right)\rightarrow Spec\left(\frac{C[y_1,\ldots,y_n]}{(\sigma^{-1}f_1,\ldots,\sigma^{-1}f_m)}\right)$ is an isomorphism between the $C$-scheme $Spec\left(\frac{C[y_1,\ldots,y_n]}{(f_1,\ldots,f_m)}\right)\rightarrow Spec(C)\xrightarrow{\sigma}Spec(C)$ and the $C$-scheme $ Spec\left(\frac{C[y_1,\ldots,y_n]}{(\sigma^{-1}f_1,\ldots,\sigma^{-1}f_m)}\right)\rightarrow Spec(C)$, that is, between $V(f_1,\ldots,f_n)^\sigma$ and $V(\sigma^{-1}f_1,\ldots,\sigma^{-1}f_m)$.
I think that the projective case should work the same as this or it could be deduced from this by glueging.
A: You can view it in this way: think of $\sigma$ as $\sigma:C'\to C$ and forget that $C'=C$, then $X^\sigma\rightarrow\operatorname{Spec}(C)\xrightarrow{\sigma}\operatorname{Spec}(C')$ is a $C'$ scheme. Then the following diagram is cartesian by easily checking the universal property
\begin{array}[c]{ccc}
X^\sigma&=&X\\
\downarrow&\square &\downarrow\\
\operatorname{Spec}(C')&\xrightarrow{\operatorname{Spec}(\sigma^{-1})}& \operatorname{Spec}(C)
\end{array}
It's well known that, if $X=\operatorname{Proj} A$ as a $C$ scheme, then $X^\sigma=X\times_{\operatorname{Spec}(C)}\operatorname{Spec}(C')\simeq\operatorname{Proj}(A\otimes_CC')$ canonically as a $C'$ scheme. So, if
$$A=C[x_0,\ldots,x_n]/(f_1,\ldots,f_m)$$
then
$$A\otimes_CC'=C'[x_0,\ldots,x_n]/(\sigma^{-1}(f_1),\ldots,\sigma^{-1}(f_m))$$
Oh, but now I remember: $C'=C$!
