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 , 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?