The proof of Belyi theorem by Lando and Zvonkin I'm sorry for asking such a specific question, but i have trouble understanding one detail in the proof of Belyi's theorem in the book "Graphs on surfaces and their applications" by Lando and Zvonkin" and i don't know how to figure it out.
I'm referring to page 149 where they want to prove the implication "existence of Belyi function" $\Rightarrow$ "definibility over $\overline{\mathbb{Q}}$". 
They take $(X,f)$ Belyi pair and $\sigma \in Gal(\mathbb{C},\overline{\mathbb{Q}})$. The Belyi pair $(X^\sigma,f^\sigma)$ is isomorphic to $(X,f)$ because the field of moduli of $(X,f)$ is a number field, so there must be an isomorphism $u:(X,f)\rightarrow (X^\sigma,f^\sigma)$.
They call $\mathcal{T}(X,f)$ and $\mathcal{T}(X^\sigma,f^\sigma)$ respectively the sets of the transcendental numbers in the coefficients of the equations defining $(X,f)$ and $(X^\sigma,f^\sigma)$.
The point i'm missing is why $\mathcal{T}(u)$, i.e. set of the transcendental numbers in the coefficients of the equations defining $u$ is an algebraic extension of the field generated by $\overline{\mathbb{Q}}\cup\mathcal{T}(X^\sigma,f^\sigma)\cup\mathcal{T}(X,f)$.
They say "because the number of such isomorphisms is finite because it doesn't exceed the number of automorphism of the dessin", but i absolutely don't understand this sentence because they define the group of automorphisms of a dessin as the centralizer of the corresponding 3-constellation in $S_n$ and  to me it doesn't seem to mean anything in this contest..
 A: The idea is that if you fix $X$, $f$, $\sigma$, and one such isomorphism $u_0$, then for any other isomorphism $u:(X,f)\to (X^{\sigma},f^{\sigma})$ it follows that the composition $u^{-1}\circ u_0$ is an automorphism of $(X,f)$.  I don't know how Lando and Zvonkin define things, but anyway, an automorphism of $(X,f)$ is usually defined to be an automorphism $\rho$ of $X$ for which $f\circ\rho = f$.  In particular, if $X$ has genus at least $2$ then it only has finitely many automorphisms.  If $X$ has genus $1$ then it has infinitely many automorphisms, but it has at most $24$ automorphisms which fix any prescribed point; since the condition $f\circ\rho=f$ implies that $\rho$ permutes the set of ramification points (a.k.a. critical points) of $f$, and since this set is finite, it follows that there are only finitely many choices for $\rho$ in this case as well.  The same argument applies when $X$ has genus $0$ if there are at least three ramification points.  It does not apply if there are exactly two ramification points, which means that $(X,f)$ is obtained from the map $z\mapsto z^n$ from the Riemann sphere to itself by composing on both sides with M\"obius transformations.  In that case $(X,f)$ has infinitely many automorphisms (but of course it is already defined over $\mathbf{Q}$).
Now that we know there are only finitely many choices for $\rho$, it follows that $\rho_0$ must be defined over a finite extension of the field $K$ over which $X$, $X^{\sigma}$, $f$, and $f^{\sigma}$ are defined.  For, suppose otherwise and let $L$ be the field of definition of $\rho$.  Then there would be infinitely many embeddings $L\to\mathbf{C}$ which are the identity on $K$, and applying any such embedding to the coefficients of $\rho_0$ yields another isomorphism $\rho: (X,f)\to (X^\sigma,f^\sigma)$.  This produces infinitely many such isomorphisms, contradiction.
