Could you explain why the restrictions of f are both group homomorphisms in Proof 2.? The rigidity lemma requires the source to be proper.

How does the action of $Aut^∘(X)$ (in this case) exhibits $X$ as a uniruled variety ?

Thanks a lot, by your answer I realize that if $X$ is covered by $U_i$, then $\#X(A)=\sum \#U_i(A)- \sum \#U_i \cap U_j(A)+\dots$ for every finite local ring $A$ as $\text{Spec} A$ is just a point so we can count points locally. But locally isomorphic does not imply global isomorphic , so we get the counterexample.

@R.vanDobbendeBruyn Oh, thanks for that observation. We can ignore bad points by base change to the good parts, but it is somehow artificial to just ignore them. Also even in this case the two curves are isogenous, so I wonder whether there will be a counter example with the proper condition or not the empty scheme over every point of $Spec \Bbb Z$.