Are all $C^1$ arcs tame? Let $n$ be a positive integer, and let $\mathbf{v}$ be a non-zero vector in $\mathbb{R}^n$.

Let $\; p : [0,1] \to \mathbb{R}^n \;$ be injective and $C^1$ and such that $p'$ is nowhere zero.

Does there always exist $\; f : ([0,1] \times \mathbb{R}^n) \to \mathbb{R}^n \;$ such that $f\hspace{0.01 in}$ is $C^1$ and for all members $t$ of $[0,1]$, $\; (\mathbf{x} \in \mathbb{R}^n) \mapsto f\hspace{0.01 in}(\langle t,\mathbf{x} \rangle) \;$ is a $C^1$ diffeomorphism of $\mathbb{R}^n$ and $\; f\hspace{0.01 in}(\langle 1,p(t)\rangle) = t\cdot \mathbf{v} \;$ ?
(That would be the $C^1$ version of an ambient isotopy.

I already know about the Fox-Artin arc, and I can see that it can 'easily' be 

made to be discontinuously differentiable with a nowhere zero derivative.)
 A: If $p$ is continuously differentiable up to and including the endpoints, then it has a $C^1$ extension to $(-\varepsilon,1+\varepsilon)$. Then it has a tubular neighborhood $U$ parametrized by a $C^1$ diffeomorphism $i:(-\varepsilon,1+\varepsilon)\times D^{n-1}\to U\subset\mathbb R^n$,
where $i(t,0)=p(t)$. In the cylinder $(-\varepsilon,1+\varepsilon)\times D^{n-1}$, it is easy to construct an explicit isotopy which shrinks the segment $[0,1]\times \{0\}$ to a small segment $[0,\delta]\times\{0\}$ and stays identical away from a small neighborhood of $[0,1]\times \{0\}$. Composing this isotopy with $i$ (and extending by the identity on $\mathbb R^n\setminus U$) you get the ambient isotopy which sends $p$ into a small sub-interval of itself.
If $\delta$ is sufficiently small, the resulting curve is contained in a graph of a $C^1$ function $f:\mathbb R\to\mathbb R^{n-1}$ (in a suitable coordinate system in $\mathbb R^n$), and it is easy to straighten it into a segment by a linear isotopy preserving the first coordinate.
