Automorphisms of surfaces Let $X$ be a projective surface with a morphism $f:X\rightarrow\mathbb{P}^1$. Assume that $f^{-1}(t)\cong\mathbb{P}^1$ for any $t\neq 0$ but $f^{-1}(0)$ is the union of two $\mathbb{P}^1$'s intersecting in one point. 
Does there exists a non trivial automorphism of $X$ restricting to an automorphism of the fiber $f^{-1}(t)$ for any $t\in\mathbb{P}^1$?
 A: I think that the answer is yes. Let me try to give a sketch of proof.
Your assumption implies that $X$ is the blow up at a point $p$ of a Hirzebruch surface $\mathbb{F}_m= \mathbb{P}(\mathcal{O}_{\mathbb P^1} \oplus \mathcal{O}_{\mathbb P^1}(m))$.
Then the automorphisms of $X$ are precisely the automorphisms of $\mathbb F_m$ fixing the point $p$. 
On the other hand, the choice of $p$ is actually irrelevant: in fact, $\textrm{Aut}(\mathbb F _m)$ acts transitively both on the unique section $C_0$ such that $C_0^2=-m$ and on its complement $\mathbb F_m \setminus C_0$, and moreover $\mathbb F_m$ blown-up in a point $p \in C_0$ is isomorphic to the blow-up of $\mathbb{F}_{m-1}$ in a point outside the negative section.  
So your question boils down whether there exist a non-trivial automorphism of $\mathbb{F}_m$ inducing the identity on the base $\mathbb{P}^1$ of the fibration $g \colon \mathbb F_m \to \mathbb P ^1$ (in fact, any such a automorphism will restrict to a non-trivial automorphism of each fibre, hence it will have lots of fixed points).
If $m=0$ then we have $\mathbb F _0 = \mathbb{P}^1 \times \mathbb{P}^1$ and then what you want is clearly true.
If $m \neq 0$, then necessarily any automorphism of $\mathbb F_m$ sends a fibre to another fibre (by the uniqueness of the fibration). Then, denoting by $\textrm{Aut}_{\mathbb P^1} (\mathbb F_m)$ the subgroup of automorphisms inducing the identity on the base, we have an exact sequence $$1 \to \textrm{Aut}_{\mathbb P^1} (\mathbb F_m) \to \textrm{Aut}(\mathbb F_m) \to \textrm{Aut} (\mathbb P^1).$$
Now for $m \geq 1$ we have $$\dim \textrm{Aut}(\mathbb F_m)=m+5, \quad \textrm{Aut} (\mathbb P^1)=3,$$
hence $\dim \textrm{Aut}_{\mathbb P^1} (\mathbb F_m) \geq m+2$, that is $\textrm{Aut}_{\mathbb P^1}(\mathbb F_m)$ is non-trivial and we are done.
