ag.algebraic geometry - Automorphism group of bi-elliptic surface

The answer is always "no". By classification, a bielliptic surface over $mathbb C$ has the form $(Etimes F)/G$ where $E,F$ are elliptic curves, $G=subset Aut(E,0)$ is an abelian group acting by complex multiplications on $E$ and by translations on $F$. ($G$ is not necessarily cyclic as Tuan correctly points out.)

($X$ maps to an elliptic curve $F/G$ and every fiber is isomorphic to an elliptic curve $E$, hence the name bielliptic.)

Then $F$ acts on $Etimes F$ by $(x,y)mapsto (x,y+f)$, and this action commutes with the $G$-action. Thus, $Fsubset Aut^0(X)$. As $F$ is a projective variety, $Aut^0(X)$ is not affine.

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