Oriented knots, equivalence relation

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.

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 $H_1\circ f_0=f_1$.