Let $k$ be an algebraically closed field.

Let $f:X\to \mathbb{P}^1$ be a smooth proper morphism with fibres of dimension one. Note that the fibres of $f$ are  geometrically connected by  Stein factorization and the fact that $\mathbb{P}^1$ is simply connected (Riemann-Hurwitz).

>>**Theorem.**  The morphism $f$ is isotrivial. 

*Proof.* Let $g$ be the genus of the fibres. Clearly, if $g=0$, then all geometric fibres are isomorphic.  Thus, if $g=0$, then $f$ is isotrivial. Next, assume that $g=1$. Then, as the moduli space of elliptic curves is affine, the moduli map associated to the Jacobian of $f$ is constant, so that $Jac(f)$ has constant $j$-invariant. It follows that $f$ is isotrivial. Finally, it is a theorem of Moret-Bailly  that any genus $g>1$ curve over $\mathbb{P}^1$ is isotrivial; see Lemme 5 in [1]. QED


[1] Laurent Moret-Bailly. *Un théorème de pureté pour les familles de courbes lisses.*  C. R. Acad. Sc. Paris, t. 300, Serie I, n. 14, 1985.

*Remark.* If $k$ has characteristic zero, then Moret-Bailly's theorem is due to Parshin. Parshin's theorem follows from the "hyperbolicity" of the moduli stack of genus $g$ ($g>1$) curves. There are many  notions of "hyperbolicity" lurking around there, and all of them imply the statement you want (in characteristic zero). (More generally, if $k$ is of characteristic zero, any smooth proper morphism $X\to \mathbb{P}^1$ whose fibres are smooth proper connected varieties with ample canonical bundle is isotrivial by work of Kovács, Migliorini, Viehweg-Zuo.) 

Let $F$ be a fibre of $f$ over a closed point of $\mathbb{P}^1_k$.

 If $g>1$, then the Isom-scheme $Isom(F\times \mathbb{P}^1, X)\to \mathbb{P}^1$ is finite etale and thus trivial. Thus, $f$ is trivial.

 

If $g=1$, then probably the family $X\to \mathbb{P}^1$ is trivial Will Sawin's argument below. 

If $g=0$, as Daniel Loughran explains in the comments below, the (isotrivial) morphism   $f$ has a section by Tsen's theorem, so that $X$ is a Hirzebruch surface.