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 $\mathbf{Sh}(X_{\mathrm{ét}})$ in the sense that adjoint functors $F_*\!:\mathbf{Sh}(X_{\mathrm{ét}})\leftrightarrows\mathbf{Sh}(X_{\mathrm{ét}}):\!F^*$ 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 $H_{\mathrm{ét}}^i(X, \mathcal{F})$. (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; see also p 89 these notes 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 "strong" assumptions on $F$ ($= F$ be universally homeom) 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 so far I know this holds 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? Invoking the quoted result above about étale site, it seems without assuming additionally universality)

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?