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The answer is always "no". By classification, a bielliptic surface over $\mathbb C$ has the form $(E\times F)/G$ where $E,F$ are elliptic curves, $G=\mathbb Z_n=(g)\subset G=\subset Aut(E,0)$ is an abelian group acting on $E\times F$ by $g.(x,y)= (g.x,y+b)$, with $nb=0$. So $G$ acts as a complex multiplication multiplications on $E$ and as a translation by translations on $F$. ($G$ is not necessarily cyclic as Tuan correctly points out.)

($X$ maps to an elliptic curve $F/\mathbb Z_n$ F/G$and every fiber is isomorphic to an elliptic curve$E$, hence the name bielliptic.) Then$F$acts on$E\times F$by$(x,y)\mapsto (x,y+f)$, and this action commutes with the$G$-action. Thus,$F\subset Aut^0(X)$. As$F$is a projective variety,$Aut^0(X)$is not affine. Post Undeleted by VA 3 added 80 characters in body; edited body; added 8 characters in body; added 7 characters in body; deleted 10 characters in body The answer is trivially always "no". For exampleBy classification, let a bielliptic surface over$X=(E\times F)/\mathbb Z_2$, \mathbb C$ has the form $(E\times F)/G$ where $E,F$ are elliptic curves, and $\mathbb Z_2$ acts G=\mathbb Z_n=(g)\subset Aut(E,0)$acting on$E\times F$by$(x,y)\mapsto g.(x,y)= (-x,y+b)$, g.x,y+b)$, with $2b=0$. nb=0$. So$G$acts as a complex multiplication on$E$and as a translation on$F$. ($X$maps to an elliptic curve$F/\mathbb Z_n$and every fiber is isomorphic to an elliptic curve$E$, hence the name bielliptic.) Then$F$acts on$E\times F$by$(x,y)\mapsto (x,y+f)$, and this action commutes with the$\mathbb Z_2$. G$-action. Thus, $F\subset Aut^0(X)$. As $F$ is a projective variety, $Aut^0(X)$ is not affine.

Note: $E\times F$ is an even easier example, but I don't think these are called (properly) bielliptic surfaces. In the classification of surfaces $E\times F$ falls in the class of abelian surfaces.

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