Let $X$ be a projective algebraic curve over some number field $K$, and let $\varphi:X\hookrightarrow \mathbb{P}^n_K$ be an embedding of it (defined over $K$) into some projective space.

Now let $X'$ be a twist of $X$; namely, a curve over $K$ that is isomorphic to $X$ over $\bar{\mathbb{Q}}$. Assume that $X'$ contains a $K$-rational point. Then does $\varphi$ induce an embedding (defined over $K$) of $X'$ into $\mathbb{P}^n_K$?

Here is a basic example I'm thinking of. Let $X$ be a smooth elliptic curve over $\mathbb{Q}$, given on an affine patch by $y^2=x^3-1$. The equation $y^2=x^3-1$ gives an embedding of $X$ into $\mathbb{P}^2_{\mathbb{Q}}$. One way to generate twists of $X$ is to look at the curves $X_a$, for $a\in\mathbb{Q}$, given on an affine patch by $ay^2=x^3-1$. The equation $ay^2=x^3-1$ gives an embedding of $X_a$ into $\mathbb{P}^2_{\mathbb{Q}}$, which, in some sense, is induced by the given embedding of $X$ into $\mathbb{P}^2_{\mathbb{Q}}$. The question is: can this be formulated in a natural way?

Here is a naive attempt. The embedding $\varphi:X\hookrightarrow \mathbb{P}^n_K$ is given by a very ample line bundle $\omega\in H^1(X,O_X^{\times})$, and a basis of global sections of $\omega$. Any twist $X'$ of $X$ is given by an element $\gamma\in H^1(K,\underline{Aut}(X))$, where $\underline{Aut}(X)$ is the sheaf of automorphisms of  $X$. A potential reformulation of my question is whether there is a natural way to "twist" $\omega$ by $\gamma$; and furthermore, if $\omega$ is very ample, and if $X'$ has a $K$-rational point, then does this imply that its twist by $\gamma$ is very ample?

If my attempt seems misguided to you, then focus on the example, which inspired my attempt to begin with...