Any algebraically closed field has tons of automorphisms. The key fact is that if $K$ and $L$ are algebraically closed and algebraic over subfields $k\subseteq K$ and $\ell\subseteq L$, then any isomorphism $f:k\to \ell$ extends to an isomorphism $\bar{f}:K\to L$. The proof is to just extend $f$ to one element of $K$ at a time: given $\alpha\in K\setminus k$, there exists $\beta\in L$ whose minimal polynomial over $\ell$ is the polynomial obtained by applying $f$ to the coefficients of minimal polynomial of $\alpha$. You can then extend $f$ to an isomorphism $k(\alpha)\to\ell(\beta)$ which sends $\alpha$ to $\beta$. Repeating this by transfinite induction, you get an extension of $f$ to an isomorphism $\bar{f}$ defined on all of $K$. The image of $\bar{f}$ must be all of $L$, since it is an algebraically closed field containing $\ell$ and $L$ is algebraic over $\ell$.
So given an algebraically closed field $K$, let $k_0$ be the prime subfield and let $B$ be a transcendence basis for $K$ over $k_0(B)$, so $K$ is algebraic over $k_0(B)$. Now let $k\subset K$ be some nontrivial finite Galois extension of $k_0(B)$ and let $f:k\to k$ be a nontrivial automorphism. By the key fact above (with $\ell=k$ and $L=K$), $f$ extends to a nontrivial automorphism of $K$.
(Note that this argument uses the axiom of choice, both in the proof of the key fact (to choose $\alpha$ and $\beta$ at every step of the induction) and in getting a transcendence basis for $K$ over $k_0$. I don't know whether it is possible for a rigid algebraically closed field to exist if you don't assume the axiom of choice.)
