Here's an argument that only trivial such automorphisms exist already for $\overline{k(x,y)}/k$ (note that in this post I don't require the automorphism to fix $\bar k$). It may be similar to the argument in the Evans-Hrushovski paper linked to by Levon Haykazyan; I haven't looked closely.
Let $\bar K / k$ have transcendence degree $\geq 2$. Say that $\phi \in Gal(\bar K / k)$ is infinitesimal if it preserves all algebraically-closed subfields. Let $G \subset Gal(\bar K / k)$ be the group of infinitesimal automorphisms. Note that $G$ is normal in $Gal(\bar K / k)$.
Lemma: If $\phi \in G$, then $\phi$ has no fixed points outside of $\bar k$.
Proof. Assume there is $\phi \in G$ with fixed points outside of $\bar k$. Then we can find $s$ a fixed point and $t$ a non-fixed point with $s,t$ algebraically independent. Let $u = \phi(t) - t$. Then
Because $\phi$ preserves $\overline{k(t)}$, we have $\overline{k(u)} \subseteq \overline{k(t)}$, since $u = \phi(t) - t$;
Because $\phi$ preserves $\overline{k(s+t)}$ we have $\overline{k(u)} \subseteq \overline{k(s+t)}$, since $u = \phi(t) - t = \phi(s+t) - (s+t)$.
Both these inclusions cannot be equalities, because $t$ and $s+t$ are algebraically indepdent and consqeuently $\overline{k(s+t)} \neq \overline{k(t)}$. But from either inclusion being proper, we conclude that $u \in \bar k$. So $\phi(t) = t + u$ with $u \in \bar k$. Let $1 < d \in \mathbb{N}$ be coprime to the characteristic of $k$. If $t^d$ is not fixed by $\phi$, then the same arguments as before show that $\phi(t^d) = t^d + u'$ with $u' \in \bar k$; if $t^d$ is fixed by $\phi$ then we still have such an equation with $u' = 0$. But we also have $\phi(t^d) = (t+u)^d$, so that $(x+u)^d - x^d - u' = 0$ is a polynomial over $\bar k$ satisfied by $t$. Because $d$ is not 1 or divisible by the characteristic of $k$ and $u \neq 0$, this is not the zero polynomial. So $t \in \bar k$, a contradiction. So there are no fixed points outside of $\bar k$.
Corollary: If $\phi \in G$ and $s \in \bar K \setminus \bar k$, then $\phi$ is uniquely determined by the value of $\phi(s)$.
Proof. If there is $\psi \in G$ with $\psi(s) = \phi(s)$, then $\psi^{-1}\phi$ fixes $s$ and so must be the identity by the lemma.
Corollary: $G$ is the group of Frobenius automorphisms, and in particular the part fixing $\bar k$ is trivial.
Proof. It suffices to show this for $\bar K = \overline{k(s,t)}$. For any automorphism $\alpha \in Gal(\overline{k(s)}/k)$, choose a lift $\beta \in Gal(\overline{k(s,t)}/k)$ which fixes $t$. If $\phi \in G$, then $\beta \phi \beta^{-1} \in G$. We have $\beta \phi \beta^{-1}(t) = \phi(t)$; by the corollary, then, $\beta \phi \beta^{-1} = \phi$. So $\phi(s) = \beta \phi \beta^{-1}(s) = \alpha \phi|_{\overline{k(s)}}\alpha^{-1}(s)$. That is, $\phi|_{\overline{k(s)}}$ lies in the center of $Gal(\overline{k(s)}/k)$, which is known to consist only of the Frobenius automorphisms. Since $\phi \in G$ is determined by $\phi|_{\overline{k(s)}}$, it follows that that $\phi$ is also a Frobenius automorphism.
Actually, I'm not sure I can prove that the center of $Gal(\overline{k(s)}/k)$ is just the Frobenius maps. So here's another argument. When we allow $s,t$ to vary, $Gal(\overline{k(s,t)}/k)$ is generated by elements like $\alpha$ above. So we see that $\phi$ in fact commutes with all of $Gal(\overline{k(s,t)}/k)$. This allows us to extend $\phi$ to an automorphism of $\bar K / k$ for any transcendence degree -- we can define it locally, and we get a homomorphism because we only have to check addition and multiplication, which are binary operations, and knowing how $\phi$ works on degree-2 extensions lets us handle this. Then we can pass to a very large $\bar K$ and conclude that because $\phi$ commutes with all automorphisms, it must be definable, i.e. definable by a polynomial. Since $\phi$ is invertible, the polynomial can only have one root, so it must be a Frobenius map.
