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The answer is no. Let $K$ be an algebraically closed field, and let $k$ be the algebraic closure of the prime field contained in $K$. Pick a transcendence basis $B$ of $K/k$, which is nonempty if $k\neq K$. Then $K$ is the algebraic closure of $k(B)$, and any automorphism of $k(B)$ extends to an automorphism of $K$. Now it is enough to show that $k(B)$ has automorphism groups with infinite orbits - for instance, note that for any $x\in B$ and $a\in k$, there is a (unique) automorphism of this field mapping $x$ to $x+a$ and fixing all other elements of $B$.