If $f'(x):\mathbb R^n \to \mathbb R^n$ is a isometry, then $f(x)=T(x)+a$. Hi folks, I'm trying to solve this one problem:
Let $f:\mathbb R^n \to \mathbb R^n$ $\in \mathrm C^1$ such that for all $x \in \mathbb R^n$, $f'(x):\mathbb R^n \to \mathbb R^n$ is a isometry (i.e. $||f'(x)\cdot v||=||v||$) with respect to the Euclidean norm.
Show that there is a Linear transformation $T:\mathbb R^n \to \mathbb R^n$ and a vector $a\in\mathbb R^n$ such that 
$$f(x)=T(x)+a, \forall x\in \mathbb R^n$$
I don't know if this information will be useful, but I've already showed that $f$ is also a isometry (i.e. $||f(x)-f(y)||=||x-y||$)
 A: According to your remark, you only need to show: 
Theorem. The only functions in $\Bbb R^d$ that are isometries are translations of orthogonal linear functions.
Proof: divided in several steps. Assume $\mathrm{A}$ is an isometry in $\Bbb R^d.$


*

*If $\mathrm{A}$ fixes the origin, then $\mathrm{A}$ preserves norm. Since $\mathrm{A}(0) = 0,$ it turns out that $\| \mathrm{A}(x) \| = \| \mathrm{A}(x) - \mathrm{A}(0) \| = \|x - 0\| = \|x\|.$

*If $\mathrm{A}$ fixes the origin, it preserves inner product; indeed, first, notice that $\|x - y\|^2 = \|\mathrm{A}(x) - \mathrm{A}(y)\|^2$ by isometry; expanding and using that $\mathrm{A}$ preserver norm (by 1.), one gets $(x \mid y) = (\mathrm{A}(x) \mid \mathrm{A}(y)).$

*If $\mathrm{A}$ fixes the origin, it preserves orthogonal basis. Indeed, let $(u_k)$ denote an orthogonal basis of $\Bbb R^d$; then $(\mathrm{A}(u_k))$ is also an orthogonal family with $d$ elements, hence a basis of $\Bbb R^d$ (this is where finiteness of the dimension is crucial in my proof; if you can take a periferical avenue and avoid this argument, the proof should go in any real Hilbert space).

*Any $x \in \Bbb R^d$ can be written as $\displaystyle x = \sum_{k = 1}^d \dfrac{(x \mid u_k)}{\| u_k \|^2} u_k$ by using its orthogonal expansion for the base $u.$ Similarly $\displaystyle \mathrm{A} x = \sum_{k = 1}^d \dfrac{(\mathrm{A} x \mid \mathrm{A} u_k)}{\| \mathrm{A} u_k \|^2} \mathrm{A} u_k = \sum_{k = 1}^d \dfrac{(x \mid u_k)}{\|u_k \|^2} \mathrm{A} u_k$ by using the expansion of $\mathrm{A} x$ with the base $\mathrm{A} u.$ So, $\mathrm{A}$ is linear.

*End of proof. The linear function $\mathrm{A}' = \mathrm{A} - \mathrm{A}(0)$ is an isometry $[\mathrm{A}'(x) - \mathrm{A}'(y) = \mathrm{A}(x) - \mathrm{A}(y)$], so it is linear. Hence, $\mathrm{A}(x) = \mathrm{T}(x) + v.$ QED

