This question is inspired by Cohomology of fibrations over the circle Moreover, it can be considered a subquestion of the above, but somehow it seems to me that some of the more interesting points were not addressed there. So I decided to ask a bit more specific question to emphasize some of those points, but I would not mind at all if someone merges this question with the above.

Let $E\to S^1$ be a fiber bundle with fiber $F$, and assume we know $H^{\bullet}(F,\mathbf{Q})$ as a ring and the monodromy action on it. Notice that since the base is a circle, the Leray spectral sequence degenerates in the second term for dimension reasons. So we have an exact sequence $0\to I\to H^{\bullet}(E,\mathbf{Q})\to Q\to 0$ where $I$ is the kernel and $Q$ is the image of $H^*(E,\mathbf{Q})\to H^{\bullet}(F,\mathbf{Q})$. In this sequence we know $Q,I$ and the action of $Q$ on $I$ from the Leray spectral sequence.

Does this suffice to determine the rational cohomology of $E$ as a ring, up to isomorphism? My guess is that probably not, but I can't find a counter-example off hand.

If not, would the higher Massey products on $H^*(F,\mathbf{Q})$ allow one to compute the cup product on $H^{\bullet}(E,\mathbf{Q})$?

If not, would a rational homotopy model $A$ of the fiber suffice, together with an automorphism $A\to A$ that covers up to homotopy the "monodromy" automorphism of the differential forms on $F$? My guess is that probably yes, but notice that computing models of fibrations with non-simply connected bases can be tricky in general: take for example the space $X$ of all (ordered) couples of distinct points of the real projective plane and the projection on the first factor: the fiber is a M\"obius band, which contracts to the circle and the monodromy action changes the sign of the generator in degree 1; but we have $H^i(X,\mathbf{Q})=\mathbf{Q}$ if $i=0,3$ and zero otherwise.