Suppose $\Omega\subset \mathbb R^2$ is a bounded domain with smooth boundary and suppose that
$$ F: \Omega \to \Omega,$$
is a diffeomorphism that fixes $\partial \Omega$ (i.e $F|\_{\partial \Omega}$ is equal to the identity map) and such that the pull back of the Euclidean metric under $F$, namely $F^\star e$ is again equal to the Euclidean metric $e$. Can we conclude that $F$ is the identity map?



On diffeomorphisms that preserve the metric
The expression "fixes $\partial\Omega$" is ambiguous. Do you mean that $f(\partial\Omega)=\partial\Omega$ or that $f(z)=z$
for all $z\in\partial\Omega$?


For the first, weaker condition, all exceptional exceptional domains $\Omega$ and functions can be easily described.


Preservation of Euclidean metric implies that $f$ is conformal, therefore it is complex analytic, and $|f'(z)|=1$
everywhere. By Complex Analysis, this easily implies that
$f(z)=cz+b,$ where $|c|=1$. Now it is easy to describe all domains $\Omega$ which can mapped homemoprphically onto themselves by such functions.


If $c$ is not a root of unity, they must be disks, including the whole plane, and any disk with center $b$ is mapped onto itself by $f(z)=cz+b-cb$. If $c$ is a root of unity
but not $1$, then $\Omega$ must have a rotational symmetry of some finite order, and again all such domains possess such homeomorphisms. Finally if $c=1$, a domain must be periodic (with period $b$).


These are all exceptional domains.


So for the second condition (fixing the boundary pointwise) we have only one exceptional domain, namely the plane, and $f(z)=cz+b$ with arbitrary $c, |c|=1$.


