The answer is "yes" if $\bar K = \overline{k(x)}$ (in fact, *every* $\phi \in Gal(\overline{k(x)}/\bar k)$ preserves all algebraically closed subfields -- there's only two of them, so it's not that hard!), but I'm not sure about higher transcendence degree. I suspect that for large enough $\bar K$ the answer must be "no", probably already at $\bar K = \overline{k(x,y)}$.


To clarify, as LSpice points out below, there are plenty of automorphisms $\phi \in Gal(\overline{k(x,y)}/\bar k)$ which do *not* preserve all algebraically closed subfields -- e.g. automorphisms swapping $x$ and $y$. So it's certainly not the case that *all* $\phi \in Gal(\bar K / \bar k)$ preserve all algebraically closed subfields. The question is for what $\bar K, \bar k$ there exists *some* such $\phi$ (other than the identity). 


For context, we might perturb the question. There are no nontrivial automorphisms of $k(x,y)/k$ preserving all subfields. Any automorphisms of a vector space preserving all subspaces, is multiplication by a scalar.


I'd really like to ask this question for an arbitrary first-order theory, but perhaps I will save that for a future post.



Can a nontrivial $\phi \in Gal(\bar K /\bar k )$ preserve all algebraically closed subfields?
Let me rephrase your question using a little bit of model theory (actually combinatorial geometry). Since the theory of algebraically closed fields is strongly minimal, the algebraic closure induces a pregeometry. More precisely given $\bar k \subseteq \bar K$ there is a pregeometry $G(\bar K/\bar k) = (\bar K, cl\_{\bar k})$ associated to it where $cl\_{\bar k}(A)$ is the algebraic closure of the field generated by $\bar k$ and $A$. Now your question becomes if there is a nontrivial automorphism of $\bar K/\bar k$ that fixes setwise all closed sets in $G(\bar K/\bar k)$.


Since we are interested only in the closed sets the right object to look at is the *geometry* $\underline G$ associated to the *pregeometry* $G = (A, cl)$. To obtain this we define an equivalence relation on $A$ as $a \sim b$ if $cl(a) = cl(b)$ and look at $(A \setminus cl(\emptyset)) / \sim$. The closure is defined as $cl(B/\sim) = \{a/\sim : a \in cl(B)\}$. Now an automorphism of $G$ induces an automorphism of $\underline G$ and it preserves closed sets of $G$ if and only if the induced automorphism on $\underline G$ is the identity.


To sum up we associate a geometry $\underline G(\bar K/\bar k)$ to the pair $\bar k \subseteq \bar K$. There is a map $\sigma : Aut(\bar K/\bar k) \to Aut(\underline G(\bar K/\bar k))$ and your questions is whether it is injective. This map actually factors through $Aut(\bar K)\_{\{\bar k\}}$ - the set of automorphisms of $\bar K$ that fix $\bar k$ *setwise*. So let $\chi : Aut(\bar K)\_{\{\bar k\}} \to Aut(\underline G(\bar K/\bar k))$ be the natural map, (i.e. $\chi(g)(cl\_{\bar k}(x)) = cl\_{\bar k}(gx)$) and $i : Aut(\bar K/\bar k) \to Aut(\bar K)\_{\bar k}$ be the inclusion map. Then $\sigma = \chi \circ i$.


The map $\chi$ is studied by Evans and Hrushovski in [The Automorphism Group of the Combinatorial Geometry of an Algebraically Closed Field](http://onlinelibrary.wiley.com/doi/10.1112/jlms/52.2.209/abstract) (Journal of LMS, Volume 52, 1995) under the assumption that the transcendence degree of $\bar K$ over $\bar k$ is at least $5$. One of their main results is that it is surjective. On page 214 they also remark that if the characteristic of $\bar K$ is nonzero, then the kernel of $\chi$ consists of powers of Frobenius and otherwise it is trivial. This then implies that the kernel of $\sigma$ is trivial. Therefore the answer to your question is negative if the transcendence degree of $\bar K$ over $\bar k$ is at least $5$.


