I am trying to show that every isometry of $(\mathbb{R}^n, g)$, regarded as a Riemannian manifold with the canonical metric $g = \delta_{ij}dx^idx^j$, is a composition of a rotation and a translation.
Here the definition of isometry is as follows:
Let $(M, g)$ and $(N, h)$ be Riemannian manifolds. An isometry from $(M, g)$ to $(N, h)$ is a diffeomorphism $\phi: M \rightarrow N$ such that $\phi^*h = g$.
My attempt:
Let $\phi: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be an isometry. Write $y = \phi(x) = (\phi^1(x), ..., \phi^n(x))$. Let $h = \delta_{kl}dy^kdy^l$. Then $$\phi^*h = (\delta_{kl}\circ\phi)\phi^*(dy^k)\otimes \phi^*(dy^l)=(\delta_{kl}\circ\phi)\frac{\partial\phi^k}{\partial x^i}\frac{\partial\phi^l}{\partial x^j}dx^idx^j = \frac{\partial\phi^k}{\partial x^i}\frac{\partial\phi^k}{\partial x^j}dx^idx^j.$$
The idea is then to equate this with $g = \delta_{ij}dx^idx^j$, which will give me quadratic equations in $\frac{\partial\phi^k}{\partial x^i}, i, k = 1, ..., n.$ After that I don't know how to proceed. It would be nice if I know that $\phi$ must be affine, but I don't know how to show this either.
I just began learning the Riemannian metric, so some of the above may not make sense. Any help is greatly appreciated.
