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Let $X$ be a smooth projective geometrically connected curve over a number field $k$ of genus $g\geq 2$.

Is the number of twists of $X$ always infinite? (The answer is no, because there aren't any twists of $X$ if the automorphism group of $X_{\bar K}$ is trivial. To obtain an example of such a curve consider a general curve.)

Let $L/K$ be a finite field extension. Are there only finitely many twists $Y$ of $X$ defined over $K$ such that $Y_L=X_L$?

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Fact 1 (The Hurwitz Bound): If $X$ is a smooth projective connected curve of genus $g\ge 2$ over $\mathbf{C}$ then $$| Aut_{\mathbf C }(X)| \le 84(g-1)$$

Fact 2: $Aut_\mathbf{C}(X) = Aut_{\overline K}(X)$ (The sentence that says "If $\phi$ is an automorphism of $X$ then $\phi$ must be one of these possibilites" is first order and thus by the first order completeness of algebraically closed fields of characteristic zero...)

Fact 3 (See Silverman's Arithmetic of Elliptic Curves chapter 10 or Serre's Galois Cohomology or Berhuy's notes or...): The twists $W_{/K}$ of a variety $V_{/K}$ are given up to isomorphism by the pointed set $H^1(Gal(\overline K /K),Aut_{\overline K}(V))$. Assume now that $L/K$ is Galois. If not you can just replace $L$ by its Galois closure. The twists which resolve over $L$ are given up to isomorphism by $H^1(Gal(L/K),Aut_{\overline K}(V))$

Fact 4 (Exercise): $$ |H^1(Gal(L/K),Aut_{\overline K}(V))| \le 84(g-1) | Gal(L/K)| $$

I must say however that I'm not sure what sections have to do with anything.

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    $\begingroup$ This is a rather late comment, but I think there's a bit more to say for Fact 3. In particular, you have written $H^1(\text{Gal}(L/K),\text{Aut}_{\overline K}(V))$, which does not make sense, since $\text{Gal}(L/K)$ does not act on $\text{Aut}_{\overline K}(V)$. Actually, you never defined $L$. I guess if you define $L$ to be Galois extension such that $\text{Aut}_{\overline K}(V)=\text{Aut}_L(V)$, which can be done due to Fact 1 (the finiteness of Aut$(V)$), it should be okay. Although in Fact 3, you didn't assume Aut$(V)$ is finite. $\endgroup$
    – Joe Silverman
    Commented Jun 5, 2017 at 18:13

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