Let $S \subset \mathbb{R}^3$ a connected surface, $p \in S$ and let $f,g:S \rightarrow S$ be two isometries.
Suppose that $f(p)=g(p)$ and $d_p f (X)= d_p g(X)$, for all $X \in T_pS$. I want to proof that $f=g$.
My guess: If we define $h:S \rightarrow \mathbb{R}^3$ with $h(x)=f(x)-g(x)$ and show that $d_x h =0$ for all $x \in S$, the result follows by connectedness of $S$. For that, we need $d_x f= d_xg$ for all $x \in S$ (and probably, here we use that $f$ and $g$ are isometries), but I do not know how to show it.
Thanks in advance!
Consider $h=g^{-1}\circ f$, $h$ is an isometry and $h$ fixes $p$ and $dh$ fixes the tangent plane $T_pS$, that is, $dh=id$.
Consider a small disk neighborhood U of p such that every point $x\in U$ is connected with p by a geodesic. Suppose $h|_U\ne id$, there is a point $x\in U$, such that $x\ne h(x)=y\in U$. As an isometry $h$ maps geodesics to geodesics, so the geodesic connecting p and x is mapped to a geodesic connecting p and y. That is $h(\exp(tX))=\exp(tY)$ which implies $dh(X)=Y$ implies $X=Y$ a contradiction.
So $h|_U=id$, so the set W={$x\in S|h(x)=x\ and\ dh_x=id$} is open. W is obviously close and non-empty thus W=S since S is connected.