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$\kappa$ must be an affine isometry. If $\gamma:[0,1]\to\mathbb{R}^n$ is a smooth curve and $L(\gamma)$ denotes its length, then $$ L(\kappa\circ\gamma)=\int_0^1|D(\kappa\circ\gamma)(t)|\, dt= \int_0^1|D\kappa(\gamma(t))\gamma'(t)|\, dt=\int_0^1|\gamma'(t)|\, dt=L(\gamma). $$ This is because $D\kappa\in O(n)$ and hence this linear map preserves lengths of vectors. The above calculation shows that $\kappa$ preserves lengths of curves from which it easily follows that it preserves distances and hence it is an isometry of $\mathbb{R}^n$. To see that $\kappa$ preserves distamces we argue as follows: If $\gamma$ is a parametrization of a segment connecting $x$ to $y$, then $\kappa\circ\gamma$ connects $x$ to $y$ and hence its length is at least $|x-y|$ $$ |\kappa(x)-\kappa(y)|\leq L(\kappa\circ\gamma)=L(\gamma)=|x-y|. $$ Applying the same argument to the inverse diffeomorphism $\kappa^{-1}$ we have $$ |x-y|=|\kappa^{-1}(\kappa(x))-\kappa^{-1}(\kappa(y))|\leq |\kappa(x)-\kappa(y)| $$ and the above two inequalities show that $|x-y|=|\kappa(x)-\kappa(y)|$.

$\kappa$ must be an affine isometry. If $\gamma:[0,1]\to\mathbb{R}^n$ is a smooth curve and $L(\gamma)$ denotes its length, then $$ L(\kappa\circ\gamma)=\int_0^1|D(\kappa\circ\gamma)(t)|\, dt= \int_0^1|D\kappa(\gamma(t))\gamma'(t)|\, dt=\int_0^1|\gamma'(t)|\, dt=L(\gamma). $$ This is because $D\kappa\in O(n)$ and hence this linear map preserves lengths of vectors. The above calculation shows that $\kappa$ preserves lengths of curves from which it easily follows that it preserves distances and hence it is an isometry of $\mathbb{R}^n$. To see that $\kappa$ preserves distamces we argue as follows: If $\gamma$ is a parametrization of a segment connecting $x$ to $y$, then $\kappa\circ\gamma$ connects $\kappa(x)$ to $\kappa(y)$ and hence its length is at least $|\kappa(x)-\kappa(y)|$ $$ |\kappa(x)-\kappa(y)|\leq L(\kappa\circ\gamma)=L(\gamma)=|x-y|. $$ Applying the same argument to the inverse diffeomorphism $\kappa^{-1}$ we have $$ |x-y|=|\kappa^{-1}(\kappa(x))-\kappa^{-1}(\kappa(y))|\leq |\kappa(x)-\kappa(y)| $$ and the above two inequalities show that $|x-y|=|\kappa(x)-\kappa(y)|$.

$\kappa$ must be an affine isometry. If $\gamma:[0,1]\to\mathbb{R}^n$ is a smooth curve and $L(\gamma)$ denotes its length, then $$ L(\kappa\circ\gamma)=\int_0^1|D(\kappa\circ\gamma)(t)|\, dt= \int_0^1|D\kappa(\gamma(t))\gamma'(t)|\, dt=\int_0^1|\gamma'(t)|\, dt=L(\gamma). $$ This is because $D\kappa\in O(n)$ and hence this linear map preserves lengths of vectors. The above caclulation shows that $\kappa$ preserves lengths of curves from which it easily follows that it preserves distances and hence it is an isometry of $\mathbb{R}^n$.

$\kappa$ must be an affine isometry. If $\gamma:[0,1]\to\mathbb{R}^n$ is a smooth curve and $L(\gamma)$ denotes its length, then $$ L(\kappa\circ\gamma)=\int_0^1|D(\kappa\circ\gamma)(t)|\, dt= \int_0^1|D\kappa(\gamma(t))\gamma'(t)|\, dt=\int_0^1|\gamma'(t)|\, dt=L(\gamma). $$ This is because $D\kappa\in O(n)$ and hence this linear map preserves lengths of vectors. The above calculation shows that $\kappa$ preserves lengths of curves from which it easily follows that it preserves distances and hence it is an isometry of $\mathbb{R}^n$. To see that $\kappa$ preserves distamces we argue as follows: If $\gamma$ is a parametrization of a segment connecting $x$ to $y$, then $\kappa\circ\gamma$ connects $x$ to $y$ and hence its length is at least $|x-y|$ $$ |\kappa(x)-\kappa(y)|\leq L(\kappa\circ\gamma)=L(\gamma)=|x-y|. $$ Applying the same argument to the inverse diffeomorphism $\kappa^{-1}$ we have $$ |x-y|=|\kappa^{-1}(\kappa(x))-\kappa^{-1}(\kappa(y))|\leq |\kappa(x)-\kappa(y)| $$ and the above two inequalities show that $|x-y|=|\kappa(x)-\kappa(y)|$.

$\kappa$ must be a linear isometry. I will write a proof soon.

$\kappa$ must be an affine isometry. If $\gamma:[0,1]\to\mathbb{R}^n$ is a smooth curve and $L(\gamma)$ denotes its length, then $$ L(\kappa\circ\gamma)=\int_0^1|D(\kappa\circ\gamma)(t)|\, dt= \int_0^1|D\kappa(\gamma(t))\gamma'(t)|\, dt=\int_0^1|\gamma'(t)|\, dt=L(\gamma). $$ This is because $D\kappa\in O(n)$ and hence this linear map preserves lengths of vectors. The above caclulation shows that $\kappa$ preserves lengths of curves from which it easily follows that it preserves distances and hence it is an isometry of $\mathbb{R}^n$.