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, a certain isom) on the étale cohomology groups $H_{\text{ét}}^i(X, \mathcal{F})$ (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 R. van Dobben de Bruyn's answer to Frobenius actions on de Rham cohomology, clarify questions on a paper of Kedlaya; see also p. 89 of these notes on étale cohomology by Brian Conrad).

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

But the question is, can the assumption on $F$ be weakened from universal 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? (E.g., is it trivial, or at least controllable in appropriate sense?)

Note, that for any morphism $G: X \to Y$ between sheaves which is a homeomorphism (not necessarily universal 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 the 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 implies that just $G$ being a homeomorphism 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 which information about $F$ we need to assure that $F$ not only acts as an isomorphism of the cohomology groups, but even as the 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?

So the key issue in my concern is how the difference between the assumptions on $F$ to be universal homeomorphism" vs "just homoemorphism" affects the statement about the action by $F$ on Zariski cohomology groups.