# Twists of rational points

Let $X$ be a "nice" algebraic curve over some field $K$, say characteristic zero.

The twists of $X$, i.e., the curves $Y$ over $K$ such that $X_{\overline K} \cong Y_{\overline K}$, are in bijection with the continuous cohomology pointed set $$H^1(\mathrm{Gal}(\overline{K}/K),X).$$

I was just wondering about an analogous question for "sections" of curves.

Now, let $x\in X(K)$. Define a twist of $x$ to be a point $y\in X(K)$ such that $x_{\overline K}$ is "isomorphic" to $y_{\overline K}$. (I don't know if it makes sense to talk about "isomorphic" sections. But what I mean is that there is an automorphism $\sigma$ of $\overline K$ such that $x_{\overline K}\circ \sigma = y_{\overline K}$.

Q1. Are twists of $x$ in bijection with some "cohomology set"?

Q2. Do there even exist any non-trivial twists of $x$? I get a feeling that $x_{\overline K}\circ \sigma$ always descends to $x$ forcing $x=y$.

Galois acts trivially on $x$ if $x \in X(K)$. (Concretely, a quasiprojective curve is defined over $K$ if and only if you can find equations over $K$ for it, and the Galois action just acts coordinatewise on solutions to those equations, so that the action is trivial on $K$-points.) So the definition of twist already implies there aren't any.
You might wonder if two $K$-point in projective space (of any dimension) could become the same over a common field extension (the point being that their coordinates could differ by a scalar) it's an exercise in Chapter I or II of Silverman to show that this can't happen, and one does use a bijection with a cohomology group - one which is known by Hilbert 90 to be zero - to show it.