In the book of Differential Geometry by Kreyszig, at page 103, it is asked that
Problem 13.1: Given two twisted curves which are in a one-to-one correspondence so that at corresponding points the tangents are parallel. Prove that at corresponding points the principal normals, and the binormals are parallel.
Let we have two curves satisfying the conditions s.t one is parametrised by $\vec x(t)$ and the other is $\vec y(t)$ s.t $c(t) \vec y(t) = x(t) c(t)$ for some real-valued scalar function, then the calculations blows up, I get nothing; but we just assume $c(t) = a$ is constant, then everything is almost trivial, so how do show this when $c(t)$ is not a constant function ?
Note that, c(t) doesn't even need to be a continuous function.
A beginning: You have two curves $t\mapsto{\bf x}(t)$ and $t\mapsto{\bf y}(t)$, and it is assumed that $\langle\dot{\bf y}(t)\rangle= \langle\dot{\bf x}(t)\rangle$ for all $t$. Here $\langle\ldots\rangle$ denotes the span of the listed vectors. We therefore may write $$\dot{\bf y}(t)=\lambda(t)\,\dot{\bf x}(t)\qquad\forall\,t\tag{1}$$ for some scalar function $t\mapsto\lambda(t)$. Differentiating $(1)$ we see that $$\ddot{\bf y}(t)=\dot\lambda(t)\,\dot{\bf x}(t)+\lambda(t)\,\ddot{\bf x}(t)\qquad\forall\,t\ .\tag{2}$$ From $(1)$and $(2)$ we can infer that not only $\langle\dot{\bf y}(t)\rangle= \langle\dot{\bf x}(t)\rangle$, but we have also $$\langle\dot{\bf y}(t),\ddot{\bf y}(t)\rangle= \langle\dot{\bf x}(t),\ddot{\bf y}(t)\rangle\qquad\forall\, t\ .\tag{3}$$ From $(3)$ we now can conclude that the principal normals ${\bf n}_{\bf y}$ and ${\bf n}_{\bf x}$ are parallel, because they are lying in the same plane, and are orthogonal to the same vector $(1)$in this plane.
The binormals, being cross products $\dot{\bf y}\times{\bf n}_{\bf y}$, resp., $\dot{\bf x}\times{\bf n}_{\bf x}$ , then are parallel as well.