Classification of isometries of $\mathbb{R}^n$ The problem is trivial yet I got stuck and have a problem with justification of the fact $\text{Isom}(\mathbb{R}^n,g=\sum_{i=1}^n (dx_i)^2)=\mathbb{R}^n \rtimes O(n)$. Suppose $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is a diffeomorphis, in order to be isometry it must satysfie $f^*g=g$ which can be translated to: $$[\frac{\partial f_i}{\partial x_j}]_{i,j}^{T}[\frac{\partial f_i}{\partial x_j}]_{i,j}=I_n.$$ Hence derivative must be orthogonal in each point. Yet I don't see why that derivative must be constant? 
EDIT: I know it can be done by considering Riemannian manifold as metric space, yet for that purpose one have to develop concept of metric and integration is needed. I would like to be able to do it directly so the above conditions fallow easily from definition of isometry yet I cant finish that. Is that argument with metric really needed here? 

The result follows easily if you think in terms of geodesics. Recall that any isometry $\phi$ of a complete Riemannian manifold $M$ is uniquely determined by $\phi(o)$ and $d\phi(o)$, at some point $o \in M$. This follows because isometries map geodesics to geodesics, according to
$$
\phi(\exp_o(v)) = \exp_{\phi(o)}\left(d\phi(o)\cdot v\right)
$$ 
In particular, if two isometries $\phi$ and $\psi$ of $M$ verify $\phi(o) = \psi(o)$ and $d\phi(o) = d\psi(o)$, then $\phi \equiv \psi$.  
Now, let $M$ be $\mathbb{R}^n$ (with the Euclidean metric, which is complete) and $\phi$ an isometry. Let $t = \phi(0)$ and $R = d\phi(0)$ where $0$ is the origin. As you pointed out, $R$ is an orthogonal linear mapping of $\mathbb{R}^n$ (identified with its tangent spaces). Then, $\phi$ must coincide with the affine transformation $\psi(x) = Rx + t$, since this $\psi$ is clearly an isometry. 
-- Salem
