Let $\mathbb{H}^3$ be the hyperbolic $3$-space. Let $F\mathbb{H}^3$ be the oriented orthonormal frame bundle, and let $\mathrm{Isom}^+(\mathbb{H}^3)$ be the orientation-preserving isometry group. It is well-known that orientation-preserving isometries of $\mathbb{H}^3$ are Mobious transformations, i.e., $\mathrm{Isom}^+(\mathbb{H}^3) \cong \mathrm{PSL}_2(\mathbb{C})$. Also, we can construct the following one-to-one correspondence, fixing a point $j = (0,0,1) \in \mathbb{H}^3$. $$ \Phi : \mathrm{Isom}^+(\mathbb{H}^3) \to F\mathbb{H}^3 \ ; \ \gamma \mapsto \left( \gamma(j), \gamma_* \left( \left. {\partial \over \partial x} \right|_j, \left. {\partial \over \partial y} \right|_j, \left. {\partial \over \partial t} \right|_j \right) \right). $$ That is, we can identify $\mathrm{Isom}^+(\mathbb{H}^3)$ and $F\mathbb{H}^3$ by fixing a point.
I want to know whether we can do the same thing for general manifolds. Let $M$ be a connected oriented Riemannian manifold. Fix $(x,f) \in FM$. Is the following map a one-to-one correspondence? $$ \Phi : \mathrm{Isom}^+(M) \to FM \ ; \ \gamma \mapsto \left( \gamma(x), \gamma_*f \right). $$ Thank you.
(Edit) I found that the above map is injective. Proposition 5.22 of Lee, Introduction to Riemannian Manifolds.
(Edit) If $M$ is simply-connected and has constant sectional curvature, then the above map is surjective. We can just rotate, translate, etc.