Automorphism of ruled surfaces associated to stable vector bundles

Let $$X$$ be a compact Riemann surface, and let $$P \rightarrow X$$ be a holomorphic $$\mathbb P^1$$-bundle over $$X$$. Then we know that $$P$$ is of form $$\mathbb P(E)$$ for some vector bundle $$E \rightarrow X$$ of rank $$2$$. At the end of section 5, Grothendieck proved that we have the following exact sequence: $$1 \rightarrow \operatorname{Aut}(E)/\mathbb{C}^* \rightarrow \operatorname{Aut}(P) \rightarrow \Gamma \rightarrow 1,$$ where $$\Gamma = \{ [T] \mid [T]\text{ is isomorphism class of a line bundle }T \text{ such that } E \cong E \otimes T\}$$. Hence $$\Gamma$$ is a finite group.

My question is what will happen if $$E$$ is a stable vector bundle? Can we expect $$\operatorname{Aut} (P) = e$$, i.e. $$\Gamma = e$$?

Not necessarily, but the counter-examples are quite particular. If $$E\cong E\otimes T$$, taking determinants give $$T^{\otimes r}\cong \mathcal{O}_X$$, with $$r=\operatorname{rk}(E)$$. Let $$\pi :\tilde{X}\rightarrow X$$ be the degree $$r$$ cyclic étale covering associated to $$T$$. The isomorphism $$E\rightarrow E\otimes T$$ defines a structure of $$\pi _*\mathcal{O}_{\tilde{X} }$$-module on $$E$$; this means that $$E$$ is of the form $$\pi _*L$$ for some line bundle $$L$$ on $$\tilde{X}$$. Conversely, for any $$L\in\operatorname{Pic}(\tilde{X} )$$ the vector bundle $$E:=\pi _*L$$ satisfies $$E\cong E\otimes T$$. It is not difficult to show that $$E$$ is semi-stable, and stable if $$L$$ is general enough.