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Assume that $M$ is a Hadamard manifold, i.e., $M$ is a Riemannian manifold such that for any point $x\in M$, the map $$ \exp_x: T_x(M)\rightarrow M $$ is defined on all $T_x(M)$ and is a diffeomorphism.

Say I fix a point $x\in M$, and an orthogonal transformation $K\in O(T_p(M))$ of the tangent space. I define a transformation $T_K:M\rightarrow M$ given by $T_K(\exp_x(v))=\exp_x(Kv)$.

Is $T_K$ an isometry of $M$?

Note that $T_K(x)=x$ and $DT_K\mid_x=K$ on the tangent space, which preserves lengths of tangent vectors by construction. But for different points, say $y,T_K(y)\in M$ I wasn't able to see what the differential map $T_y(M)\rightarrow T_{T_K(y)}(M)$ is, and whether it preserves lengths.

Levent
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    This would imply that the isotropy group of $M$ at $p$ is $O(n)$. Does this help? – Laz Feb 17 '22 at 15:42
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    No, just think of the product ${\mathbb R}\times {\mathbb H}^2$. Moreover, one can find Hadamard manifolds without any symmetries (just take the hyperbolic plane and make a small perturbation of the metric on a nonempty open subset). – Moishe Kohan Feb 17 '22 at 19:54

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