Question:
Show that the knowledge of the vector function $n=n(s)$ of a curve $\alpha:I\rightarrow \mathbb{R^3}$ with nonzero torsion everywhere, determines the curvature $k(s)$ and the torsion $\tau (s)$ of $\alpha$.
Notes: $n$ is the normal versor to $\alpha$.
Attempt: I tried using Frenet-Serret formulas, and then using the vector product between $n$ and $n'$, but it seems like I can't get to any result.
The Question as posed here is not solvable (as is indicated by the comments in the answer above). One needs the knowledge of the function $\frac{\kappa}{\tau}$ at one point $t_0$.
Lets give a counterexample:
Consider the Helix
$$ c(s) := (a \cdot \cos(s) , a \cdot \sin(s) , b \cdot s) \text{ for } s \in \mathbb{R}$$
with $a^2 +b^2 = 1$, $a,b>0$. Then $c(s)$ ist parametrized by arclength.
The normal vector is given by
$$ n(s) = (- \cos(s) , -\sin(s), 0) \text{ for all } s\in \mathbb{R}. $$
One has in general $\kappa=a, \tau= -b$ and $\frac{\kappa}{\tau}=-a/b$.
The choices $a_1=b_1= 1/\sqrt{2}$ and $a_2= 1/2 ,b_2= \sqrt{3}/2$ give two different curves parametrized by arclength. Both curves have the same normal vector for all times and non-vanishing torsion. And both curves have different curvature and torsion.
Hint: This is rather a long calculation. If you need any more steps feel free to ask.
The idea though is to show that $$\frac{(n \wedge n') \cdot n''}{|n'|^2} = \frac{(\frac{\kappa}{\tau})'}{(\frac{\kappa}{\tau})^2 + 1} : = a (s)$$
Then,
$$\int a(s) ds = \arctan \left(\frac{\kappa}{\tau}\right)$$
Thus $\kappa/\tau$ can be determined, plus we have that $\kappa > 0$ then we may get the sign of $\tau$. Finally, use
$$|n'|^2 = |- \kappa t - \tau b|^2 = \kappa ^2 + \tau^2$$
to determine $\kappa^2$ and $\tau^2$.
