I will denote by $X$ the set of rational points on the unit circle, i.e., $$X := \{ (x,y) \in \mathbb{Q}^2: x^2 + y^2 = 1 \}.$$ Viewing both $\mathbb{Q}$ and $X$ as an affine varieties, then every morphism $f: \mathbb{Q} \to X$ is given by two (global) regular functions on $\mathbb{Q}$. To be more specific, $$f(t) = \left(f_1(t),f_2(t)\right)$$ such that both $f_1$ and $f_2$ are rational functions of the form $\frac gh$, where $g$ and $h$ are polynomials on $\mathbb{Q}$, and $h$ never vanishes on $\mathbb{Q}$. My question is: does there exist an isomorphism $f: \mathbb{Q} \to X$?
On one hand, I know that the rational parametrization of the unit circle $$ t \mapsto \left( \frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2} \right) $$ is a morphism from $\mathbb{Q}$ to $X$, but it's not an isomorphism. However, this doesn't mean there cannot be an isomorphism $f: \mathbb{Q} \to X$.
On the other hand, if we equip $\mathbb{Q}$ and $X$ with the usual topologies (i.e., the subspace topologies from $\mathbb{R}$ and $\mathbb{R}^2$, respectively), then $X$ is a countable metrizable space without isolated points, and hence homeomorphic to $\mathbb{Q}$, by a theorem of Sierpinski whose proof can be found here. So my question is indeed asking whether a homeomorphism $f: \mathbb{Q} \to X$ can be in the form of two rational functions $f_1$ and $f_2$ as mentioned above.
No, there is no isomorphism $\mathbb{Q}\to X$.
There is some risk of confusion here, because your notion of morphism for $\mathbb{Q}$-varieties is not the usual one (a morphism of $\mathbb{Q}$-varieties is usually required to be regular everywhere, not just on rational points). I will write $\mathbb{A}^1$ for the affine line over $\mathbb{Q}$, and $Y:=\mathbb{V}(x^2+y^2-1)\subset\mathbb{A}^2$. The $\mathbb{Q}$-rational points of $\mathbb{A}^1$ are $\mathbb{Q}$ and the $\mathbb{Q}$-rational points of $Y$ are your $X$.
In what follows, by "morphism" I mean morphism in the usual sense (so a morphism of affine varieties corresponds to a ring homomorphism going the other way). If I understand correctly, your question is whether there exists a birational morphism $f:\mathbb{A}^1\to Y$ that is bijective on $\mathbb{Q}$-points. Assume $f$ exists. Both $\mathbb{A}^1$ and $Y$ have $\mathbb{P}^1$ as a compactification, and $f$ extends to an isomorphism $f:\mathbb{P}^1\to \mathbb{P}^1$. Now $\mathbb{A}^1(\mathbb{Q})$ is a proper subset of $\mathbb{P}^1(\mathbb{Q})$, so $f(\mathbb{A}^1(\mathbb{Q}))$ must be a proper subset of $\mathbb{P}^1(\mathbb{Q})$. However, the inclusion $Y\subset\mathbb{P}^1$ is an equality on $\mathbb{Q}$-points, so we have $f(\mathbb{A}^1(\mathbb{Q}))\subsetneq Y(\mathbb{Q})$. In your language, this means $f(\mathbb{Q})\subsetneq X$.