Oriented knots, equivalence relation I've seen the following definition of oriented knots and knot equivalence:
$\textbf{Definition 1:}$ An $\textit{oriented knot}$ in $\mathbb{R}^3$ is a (topological) embedding $f:S^1\rightarrow \mathbb{R}^3$ of the circle into $3$-space.
$\textbf{Definition 2:}$ Two oriented knots $f_0,f_1:S^1\rightarrow \mathbb{R}^3$ are said to be $\textit{equivalent}$ or $\textit{ambient isotopic}$ if there exists a continuous map $H:\mathbb{R}^3\times [0,1]\rightarrow \mathbb{R}^3$, such that
$(i)$ For every fixed $t\in [0,1]$, the map $H_t:\mathbb{R}^3\rightarrow \mathbb{R}^3,$ $x\mapsto H(x,t)$ is a homeomorphism.
$(ii)$ $H_0=Id|_{\mathbb{R}^3}$
$(iii)$ $H_1\circ f_0=f_1$
Ambient isotopy defines an equivalence relation on the set of all (topological) embeddings of $S^1$ into $\mathbb{R}^3$. There is a standard orientation of $S^1$ (viewed as a subset of the Euclidean plane) and so every given embedding gives rise to an orientation of its image.
Now if the above definitions really capture the intuition behind oriented knots, then intuitively any two embeddings $f_0,f_1:S^1\rightarrow \mathbb{R}^3$ with $f_0(S^1)=f_1(S^1)$ and for which $f_1^{-1}\circ f_0:S^1\rightarrow S^1$ is an orientation preserving homeomorphism represent the same oriented knot. Hence I expect $f_0$ and $f_1$ to be contained in the same equivalence class.
Is this true in general or maybe only for tame knots? I was not able confirm my conjecture in either case. Do I miss something obvious or is this problem more involved than I suspect?
Kind regards
Dennis
 A: It's false in general that if $f_1,f_2 : S^1 \to \mathbb R^3$ are oriented knots according to Definition 1, and if there exists an orientation preserving homeomorphism $g : S^1 \to S^1$ such that $f_2 = f_1 \circ g$, then $f_1$ and $f_2$ are ambient isotopic according to Definition 2. 
For a counterexample, take a look at the picture of the wild knot on wikipedia. That knot is locally tame at every point except for one point which I'll denote $x$. So a self-ambient-isotopy of this knot cannot move the point $x$ to any other point on the knot.
A: As others have mentioned, it is not true for wild knots, since a homeomorphism of $\mathbb{R}^3$ preserving the image of a knot must also preserve the wild points of the knot (points where it cannot locally be thickened), and so if you reparametrize your knot to map different points of $S^1$ to the wild points then the required ambient isotopy cannot exist.
It is true for tame knots.  Here's the idea: if $f_1$ differs from $f_0$ by an orientation-preserving reparameterization $g$, take an isotopy $g_t$ between the identity and $g$ on $S^1$.  Now to get an ambient isotopy between $f_0$ and $f_1$, use the isotopy $g_t$ on the image of the knot, and interpolate between that and the identity on a thickening of the knot, so that then you can extend to all of $\mathbb{R}^3$ by just taking the identity outside the thickening.  
Here are the details.  Suppose $f_0:S^1\to\mathbb{R}^3$ is a tame knot, which extends to an embedding $F_0:S^1\times D^2\to\mathbb{R}^3$.  Let $g:S^1\to S^1$ be an orientation-preserving diffeomorphism and let $f_1=f_0\circ g$.  To construct an ambient isotopy between $f_0$ and $f_1$, lift $g$ to a map $G:\mathbb{R}\to\mathbb{R}$ on the universal covers (here we consider $\mathbb{R}$ as the universal cover of $S^1$ via $x\mapsto \exp(2\pi i x)$).  Since $g$ was an orientation-preserving diffeomorphism, $G$ is strictly increasing and satisfies $G(x+1)=G(x)+1$ for all $x$.  Now let $$G_t(x)=tG(x)+(1-t)x$$ and observe that $G_t$ is also increasing and satisfies $G_t(x+1)=G_t(x)+1$ for all $x$.  Thus $G_t$ descends to an orientation-preserving diffeomorphism $g_t:S^1\to S^1$ with $g_0$ the identity and $g_1=g$, and $g_t(x)$ being jointly continuous in $t$ and $x$.  Now define $$H_t(x)=F_0(g_{(1-|s|)t}(y),s)$$ if $x=F_0(y,s)$ for $(y,s)\in S^1\times D^2$ and $H_t(x)=x$ if $x$ is not in the image of $F_0$.  Observe first that $H$ is continuous, since at the boundary of the image of $F_0$, $|s|=1$ so the formula above gives $H_t(x)=F_0(g_0(y),s)=F_0(y,s)=x$ since $g_0$ is the identity.  Also, for any $t$, $H_t$ is a homeomorphism since $(y,s)\mapsto (g_{(1-|s|)t}(y),s)$ is a homeomorphism $S^1\times D^2\to S^1\times D^2$.  Finally, $H_0$ is the identity since $g_0$ is the identity and $$H_1(f_0(y))=H_1(F_0(y,0))=F_0(g_1(y),0)=f_0(g(y))=f_1(y)$$ so $H_1\circ f_0=f_1$.
