My question involves the symmetries of a (pseudo)-Riemannian metric preserving the Levi-Civita connection (LCC), its unique torsion-free metric connection. For a basic example, one notes that the standard flat connection on $\mathbb R^d$ is the LCC of $\sum_i a_i dx_i\otimes dx_i$ for any non-zero constants $a_i$. This example suggests that sufficiently rigid transformations of a metric may fix the LCC.

Question. Let $M$ be a (smooth) manifold. Let $\mathcal G$ be a maximal group of vector bundle automorphisms of $\DeclareMathOperator{Sym}{Sym}\Sym^2(T^*M)$ with the property that the action of $\mathcal G$ on the non-degenerate sections leaves the LCC invariant. What are the geometries of the $\mathcal G$-orbits? Is $\mathcal G$ finite dimensional?

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    $\begingroup$ By "leaves the LCC" invariant, do you mean that it leaves $\nabla X$ invariant as an element of $\text{Hom}(TM,TM)$ for each $X\in\Gamma(TM)$? $\endgroup$
    – S.Surace
    Apr 5 '19 at 11:28
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    $\begingroup$ Yes, meaning that for any fixed vector fields $X,Y$, for any non-degenerate metric $m$, and any element of the automorphism group $h\in \mathcal G$ described above, the torsion-free metric connections $\nabla_m$ for $m$ and $\nabla_{h\cdot m}$ for $h\cdot m$ satisfy $(\nabla_mX)(Y)=(\nabla_{h\cdot m}X)(Y)$. $\endgroup$
    – pre-kidney
    Apr 6 '19 at 20:13

I'm not sure this answers your question precisely, but here I described how many Riemannian metrics you have with prescribed Levi-Civita connection, it depends on the holonomy of the connection:



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