Let $(M,g)$ be a closed smooth Riemannian manifold, and denote by $G$ a closed subgroup of its isometry group. By considering the maps $g^*$ induced by elements $g\in G$ in the (de Rham) cohomology $H^*(M,\mathbb R)$, one obtains a natural homomorphism $\phi\colon G\to \mathrm{GL}(H^*(M,\mathbb R))$. The image of this homomorphism is a finite subgroup $\Gamma\subset \mathrm{GL}(H^*(M,\mathbb R))$.
My question is if all finite subgroups $\Gamma$ as above arise in this way (assuming no smoothness issues); more precisely:
Question: Suppose that a finite subgroup $\Gamma\subset \mathrm{GL}(H^*(M,\mathbb R))$ is the image of a subgroup of diffeomorphisms $G\subset Diff(M)$ under the homomorphism $\phi\colon Diff(M)\to \mathrm{GL}( H^*(M,\mathbb R))$, $\phi(f)=f^*$, that is, $\Gamma=\phi(G)$. Then, is it true that $\Gamma$ is also the image of a subgroup of isometries under this homomorphism? That is, can we always realize $G\subset Diff(M)$ as a subgroup $G\subset Iso(M,g)$?
Edit [See comment below by Will Sawin]: The answer is (trivially) NO, unless question is reformulated to allow $g$ to vary, and allowing $g$ to vary would lead to too vague of a question...
Remark 0: I understand this is related to the Nielsen realization problem, which is essentially the above question when $(M,g)$ is a hyperbolic surface (see wiki: https://en.wikipedia.org/wiki/Nielsen_realization_problem). I am mostly interested in the case $\dim M\geq 3$, and preferably without any curvature assumptions on $g$.
Remark 1: The reason for the hypotheses that $\Gamma=\phi(G)$ where $G\subset Diff(M)$ is to avoid any "smoothness" issues that could appear in attempting to realize automorphisms of the cohomology ring with diffeomorphisms. An example borrowed from Block and Weinberger (http://math.uchicago.edu/~shmuel/Nielsen.pdf) is the following: let $M=(T^7\# \Sigma^7)\times S^1$ be the product of an exotic 7-torus with a circle, and $\Gamma=S_8$ the permutation group acting on $H^*(M,\mathbb R)$ in the same way it would act on $H^*(T^8,\mathbb R)$ if we had not messed up the smooth structure. Then $\Gamma$ cannot be realized as $\phi(G)$ for any $G\subset Diff(M)$, let alone by isometries of a given Riemannian metric on $M$. So, in my question above, I want to bypass this type of obstruction and focus only on the "diffeo to isometry" part of the question.
Remark 2: In case the answer is affirmative, it would be neat if there was some variational procedure to construct $G\subset Iso(M,g)$, for example by taking diffeos and minimizing some kind of displacement with respect to $g$, or the ratios between the eigenvalues of its linearization.
