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[NOT homework] I am studying for my qualifying exam, and a past question is the following:

Let $(M, g)$ be a connected Riemannian manifold and $f_1, f_2 : M \to M$ be isometries. Suppose that exists a point $p$ in $M$ so that $f_1(p)=f_2(p)$ and $(df_1)_p=(df_2)_p$. Show that $f_1=f_2$

My idea is to consider the set $S = \{p \in M : f_1(p)=f_2(p)\text{ and }(df_1)_p=(df_2)_p\}$. Since $M$ is connected, if $S$ is both open and closed, then $S$ will be $\emptyset$ or $M$. But from the hypothesis, $p \in S \implies S=M \implies f_1=f_2$.

[EDIT]: Using @Travis comment and this discussion, the result follows

tomate
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    You might consider the map $\phi := f_2^{-1} \circ f_1$, which satisfies $\phi(p) = p$ and $d\phi_p = {\bf 1}_p$. – Travis Willse Jul 16 '23 at 19:51
  • Hey, @TravisWillse! Thanks for your comments. Yes, it does answer indeed. I edited the post linking to this other thread – tomate Jul 16 '23 at 22:05
  • I actually found the post via your link and then voted to mark this post as a duplicate so it wasn't hanging open. This is a classic differential geometry question, by the way! – Travis Willse Jul 17 '23 at 00:22

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