Is a bijective morphism of quasi-affine smooth varieties an isomorphism?

We have a bijective morphism $f:X\to Y$ of quasi-affine varieties (say over $\Bbb{C}$).

Can $f$ fail to be an isomorphism if $X$ and $Y$ are smooth?

Hartshorne gives the Froebenius morphism as an example (exercise I.3.2 b). If $k=\bar k$, and $\operatorname{char}(k)=p\neq 0$, then the map $\Bbb A^1_k\to \Bbb A^1_k$ given by $x\mapsto x^p$ is bijective (and even a homeomorphism), but not an isomorphism of varieties.

We might as well think that our morphism is bijective on scheme-theoretic points, i.e. is quasi-finite.

By Zariski's main theorem a quasi-finite map $X \to Y$ factors as a composition of an open immersion and a finite map, $X \to W \to Y$. A degree 0 finite morphism ($W \to Y$) with normal target, which $Y$ is since it is smooth, has to be isomorphism in char 0 (would be radicial in general): the field extension it enduces on the general fibre is trivial, since it is of degree 1.

Since the composition is surjective $X \to W$ is an isomorphism too, and we are done.

You don't actually need quasi-afineness assumption.