Let $d\in\mathbb N$ and $\Omega_i\subseteq\mathbb R^d$ be open.
If $f$ is an isometry from $\Omega_1$ to $\Omega_2$, can we conclude that $f$ is Fréchet differentiable and $\left|\det{\rm D}f\right|=1$? On the other hand, if $g$ is a $C^1$-diffeomorphism from $\Omega_1$ onto $\Omega_2$ with $\left|\det{\rm D}g\right|=1$, does $g$ need to be an isometry?
I'm unsure how to approach this. I know that if $h:\mathbb R^d\to\mathbb R^d$ is a linear isometry, then it is clearly Fréchet differentiable with ${\rm D}h=h$. Moreover, $h$ is orthogonal and hence $\left|\det{\rm D}h\right|=\left|\det h\right|=1$.