Let $X$ and $Y$ be compact manifolds. $p:X \rightarrow Y$ is a covering. Generally it is not true that
$$ H^* (X , \mathbb{Q} ) = H^* (Y , \mathbb{Q} )$$
For instance, if $Y$ is a sphere with $y$ handles (let $y > 1$), then $X$ is a sphere with $x$ handles. Moreover if it is not trivial covering then $x > y$.
But if $X$ and $Y$ are Lie groups and $p$ is homomorphism, then $ H^* (X , \mathbb{Q} ) = H^* (Y , \mathbb{Q} )$. It seems a bit like a miracle.
To prove it you have to consider Chevalley-Eilenberg complex. It counts the cohomology of Lie group. On the other hand it is formulated purely in terms of Lie algebra.
Question: I want a purely topological condition on $p$, $X$ and $Y$ to keep isomorphism $$ H^* (X , \mathbb{Q} ) = H^* (Y , \mathbb{Q} )$$.
For instance, what about condition "$Y$ has trivial tangent bundle"?
If $X \to Y$ is a finite Galois cover with Galois group $G$ then it's known that the induced map $H^{\bullet}(Y, \mathbb{Q}) \to H^{\bullet}(X, \mathbb{Q})$ is injective and induces an isomorphism
$$H^{\bullet}(Y, \mathbb{Q}) \cong H^{\bullet}(X, \mathbb{Q})^G.$$
(This has nothing to do with $X$ and $Y$ being compact or manifolds.) So here the answer is that the natural map is an isomorphism iff the action of $G$ on $H^{\bullet}(X, \mathbb{Q})$ is trivial. (This has nothing to do with the tangent bundle of $X$ or $Y$.)
If $p : X \to Y$ is a finite covering map of connected Lie groups (both "finite" and "connected" are essential to what follows; I leave it to you to find counterexamples if either is dropped), then it is Galois with Galois group $G = \text{ker}(p)$, and the covering map is given by quotienting by $G$. The action of $G$ on $H^{\bullet}(X, \mathbb{Q})$ is trivial because it factors through the action of $X$, which is trivial because $X$ is connected. No need to pass to Lie algebras.
So you can expect the action of $G$ to be trivial if it factors through the action of a connected Lie group; for example, you might know that $G$ acts by isometries of some Riemannian metric on $X$ with respect to which the isometry group of $X$ is a connected Lie group. Other than that, all bets are off.
Edit: Note also that if $X, Y$ are compact manifolds then they have Euler characteristics which satisfy $\chi(X) = n \chi(Y)$ where $n$ is the degree of the cover, so a necessary condition in this case (if $n \neq 1$) is that $\chi(X) = \chi(Y) = 0$, which is of course always satisfied for compact Lie groups so no problem there. The way this is reflected in the above argument is that if $G$ is a nontrivial group acting freely on $X$ but trivially on the cohomology of $X$, then by the Lefschetz fixed point theorem $\chi(X) = 0$.