Conformal transformation on Riemannian manifold with boundary For a conformal transformation on an $n$-dimensional ($n>2$) Riemannian manifold with boundary. If the transformation becomes an identity on a portion of the boundary, can one conclude that the transformation is actually an identity map?
 A: Lichnerowicz's conjecture (Ferrand's theorem) states that, with two exceptions (manifolds conformal to the round sphere and the Euclidean space), if $(M,h)$ is a connected Riemannian manifold and $G=Conf(M,h)$ is the group of conformal automorphisms, then there exists a smooth positive function $\rho$  on $M$ such that $G< Isom(M, \rho h)$, where $Isom$ denotes the group of isometries. I did not check, but it very likely that the same proof works for manifolds with boundary.
There will be few more exceptions but they all will be conformally flat and, hence, the uniqueness you are asking for will also hold in the exceptional case since we will be dealing with subgroups of the group of Moebius transformations.
Ferrand, Jacqueline, The action of conformal transformations on a Riemannian manifold, Math. Ann. 304, No. 2, 277-291 (1996). ZBL0866.53027.
From your comment, it is clear that you know how to finish the proof in the case of isometric maps (it is indeed quite easy).
Remark. 1. Ferrand's theorem has complicated history that Ferrand discusses in the introduction to her paper. Briefly: Ferrand gave a sketch of a proof for compact manifolds in 1971; in 1972 Alexeevsky gave a proof in the general case. It turned out later on that his proof was incorrect. Ferrand extended her argument from the compact case to general manifolds in 1996.


*For the uniqueness theorem to hold, you need the manifold to be connected and a "piece" of the boundary to mean an open nonempty subset of the boundary.

Edit. I checked the argument in the case you mentioned in a comment, namely, when $(M,h)$ is a compact Riemannian manifold with smooth boundary. Suppose that $f\in Conf(M,h)$ and $f$ fixes 3 distinct points on the boundary of $M$. Then the subgroup of $Conf(M,h)$ generated by $f$ is relatively compact in $Conf(M,h)$ due to the convergence property of $Conf(M,h)$ and the latter holds regardless of whether $M$ has boundary or not (see the relevant discussion in Ferrand's paper, section 7). Once you have relative compactness, you will find an $f$-invariant conformal metric on $(M,h)$ (of the form $\rho h$) by averaging.
