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 (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 $5$.