Every such automorphism is contained in the automorphism group of the field or rational functions $F(t)$ over $F$, which equals $\mathrm{PGL}_2(F)$, and so is a finite group.
[Edit:] upon further reflection, my answer is incomplete, because $\mathrm{PGL}_2(F)$ is the group of isomorphisms of $F(t)$ as an $F$-algebra, not as a ring. Call $G$ the group of automorphisms of $F(t)$ as a ring. Since $F$ is the algebraic closure of the prime field $\mathbb F_p$ in $F(t)$, every element of $G$ induces an automorphism of $F$; this induces a homomorphism of $G$ onto the automorphism group of $F$, which is finite, with kernel $\mathrm{PGL}_2(F)$. Both groups are finite, so $G$ is finite.