If there is a point $p \in M$ such that $f(p) = g(p)$ and $df_p = dg_p$ then $f = g$. Let $M,N$ Riemannian manifold conected and $f,g:M \rightarrow N$ two isometries. If there is a point $p \in M$ such that $f(p) = g(p)$ and $df_p = dg_p$ then $f = g$. 
Comments: 
I'm considering the set $A = \{q \in M ; f(q) = g(q) \ \text{and} \ df_q = dg_q\}$. 
The set is not empty and closed, but I can not show it is open.

Consider $x\in A$, there exists a neigborhood $U$ of $A$ such that $y\in U$ implies that $y=\exp_x(v), v\in T_xM$, $f(\exp_x(v))=\exp_{f(x)}(df_x(v))=\exp_{g(x)}(dg_x(v))$ since $f$ and $g$ are isometries, where $\exp_x(v)$ is defined as follows: let $c(t)$ be the geodesic such that $c(0)=x$ and $c'(0)=v, \exp_x(v)=c(1)$. This implies that the restriction of $f$ and $g$ to $U$ are equal and $U$ is open.
