Let $F: X \to X$ to be an endomorphism of scheme $X$, which is additionally assumed to induce an *universal homeomorphism* on the underlying topological space $| X|$. Then it is known that this induces an selfequivalence of the etale topos $\mathbf{Sh}(X_{\text{ét}})$ in the sense that adjoint functors $F_*\!:\mathbf{Sh}(X_{\text{ét}})\leftrightarrows\mathbf{Sh}(X_{\text{ét}}):\!F^*$ are even inverse. 

Now it is known that this has as consequence that induced $F^*$ acts trivially ($=$ literally as identity, and *not* as say certain isom) on the etale cohomology groups $H_{ét}^i(X, \mathcal{F})$. (SGA 5, Exp. XV, §2, Prop. 2(c); the idea is also elaborated for case $F$ beeing the absolute Frobenius by Remy van Dobben de Bruyn in comments below [this answer][1]; see also p 89 [these notes][2] by Brian Conrad)


**Question:** What can be said about the induced action by $F$ on Zariski cohomology $H^i_{zar}(X, -)$ instead of etale if we weaken the assumption on $F$ from *universally homeomorphism* to *only* homeomorphism?  
As etale topos is finer then Zariski topos, under above assumptions on $F$ the equivalence by $F^*$ above for etale topos implies equiv on Zariski topoi, so once $F$ is universally homeom, identical argument shows that $F$ acts also on Zariski cohomology trivially.

But the question is, can the assumption on $F$ be weakened from *universally homeomorphism* with the *"price"* we are ready to pay be that we haven't any more a statement as above for triviality of action by $F$ on étale cohomology, but maybe passing to coarser topology we can say something interesting about induced action by $F$ on Zariski cohomology? (eg is it trivial, or at least controlable in appropriate sense?)


Note, that for any morphism $G: X \to Y$ between sheaves which is a homeomorphism (*but neccessarily universally homeo(!)*), that $G_*$ still gives rise to  equivalence of cats $\mathbf{Sh}(X_{\text{zar}})\to \mathbf{Sh}(Y_{\text{zar}})$ (a priori only for Zariski site if I'm not confusing something; see somewhere in Mac Lane and Moerdijk's "Sheaves in Logic")  
 (...btw is this something "Zariski site specific" or does it hold also for reasonable refinements , eg étale?)


Especially such equivalence imply that only $G$ homoemorphism alone ( especially a pure *"topological"* property, although one should still keep in mind that $G$ has to be a schematic morphism; otherwise refined constructions could steer into serious troubles...) suffice to assure that already the cohomology groups $H^i(X, \mathcal{F})$ and $H^i(Y, G_* \mathcal{F})$ are somehow isomorphic.

But coming back to case $Y=X$ and $G=F$: The question is whichinformation about $F$ we need to assure that $F$ not only acts as isomorphism of the cohomology groups, but even as identity? 

Above on étale site it suffice to require that $F$ is universal homeo, and not just homeo. But if we restrict to coarser Zariski site, how far can the assumption on $F$ to be universal homeom to be weakened to still have that $F$ acts as identity on cohomology groups?

  [1]: https://mathoverflow.net/a/299575/108274
  [2]: http://math.stanford.edu/~conrad/Weil2seminar/Notes/etnotes.pdf