Let $\gamma(t): [0, 1] \rightarrow \mathbb{R}^{2}$ be a counterclockwise-oriented closed loop that does not intersect itself and does not intersect the origin: $\gamma(0)=\gamma(1)$. Further, assume that the winding number of this curve about the origin is 1. Does there exist an ambient isotopy of $\mathbb{R}^{2}$ that takes $\gamma(t)$ to $(\cos t, \sin t)$? I believe a homeomorphism from $\gamma(t)$ to $(\cos t, \sin t)$ can be constructed by sending an arbitrary nonzero point $(x, y)$ to the the point $\frac{1}{|\gamma(T_{xy})|}(x, y)$ where $\gamma(T_{xy})$ is the point where $\gamma(t)$ intersects the ray from $(0, 0)$ to $(x, y)$. Call this homeomorphism $H_{1}$. This assumes that that $\gamma(t)$ only intersects the ray from $(0, 0)$ to $(x, y)$ at exactly one point. In this case, an ambient isotopy of $\mathbb{R}^{2}$ taking the identity to $H_{1}$ (and hence $\gamma(t)$ to $(\cos t, \sin t)$ ) can be constructed as follows $H_{t}(x, y) = \frac{(1-t)|\gamma(T_{xy})| + t}{|\gamma(T_{xy})}(x , y)$. Each $H_{t}$ should be a homeomorphism because the factor $\frac{(1-t)|\gamma(T_{xy})|+t}{|\gamma(T_{xy}|}$ is never zero when $t$ is between $0$ and $1$. This also presumes that each homeomorphism sends $(0,0)$ to $(0,0)$.
However, an arbitrary non-intersecting closed curve $\gamma(t): [0, 1] \rightarrow \mathbb{R}^{2}$ (even one that has winding number 1), may intersect a ray from $(0,0)$ to a nonzero $(x, y)$ more than once. How can the above ambient isotopy of $\mathbb{R}^{2}$ be extended to one that takes an arbitrary non-intersecting closed loop, $\gamma(t)$, of winding number 1, to $(\cos t, \sin t)$. Is it possible to first construct a homeomorphism, $H_{\frac{1}{2}}$ of $\mathbb{R}^{2}$ that takes $\gamma(t)$ to another closed loop that only intersects the ray from $(0,0)$ to $(x, y)$ at exactly one point. This homeomorphism could then be extended to an isotopy from the identity to $H_{\frac{1}{2}}$. The above argument would be then be used to construct an isotopy from $H_{\frac{1}{2}}$ to $H_{1}$. However, I am unsure how to fill in the details of constructing the homeomorphism $H_{\frac{1}{2}}$ or the isotopy between it and the identity. Any assistance here would be greatly appreciated. Thanks!
Here's a proof using the Schönflies Theorem together with the Classification of Surfaces.
The first step is to apply the Schönflies Theorem, which gives you a homeomorphism of the plane $h : \mathbb R^2 \to \mathbb R^2$ such that $\gamma' = h \circ \gamma$ is a counterclockwise parameterization of the unit circle. Your original counterclockwise parameterization of the unit circle, let me call it $\delta$, has been taken to another simple closed curve $\delta' = h \circ \delta$.
The second step is to radially isotope $\gamma'$ to a circle that I'll denote $\gamma''$, and which is so large that $\delta'$ lies on the inside of $\gamma''$.
The third step is to consider the closure of the region lying on the inside of $\gamma''$ and the outside of $\delta'$; let's call this $A$. It follows that $A$ is a compact surface with boundary, the only issue here being to check that each component of the boundary has a collar neighborhood. This is obvious for the circle $\gamma''$, using an inner coller neighborhood. And then the outer collar neighborhood of $\delta$ provides a collar neighborhood of $\delta'$ in $A$.
The fourth step is to apply the classification of surfaces to $A$, to conclude that $A$ is homeomorphic to the annulus $S^1 \times [0,1]$ (I can supply more details here as needed), with $S^1 \times \{0\}$ corresponding to $\gamma''$ and $S^1 \times \{1\}$ correpsonding to $\delta'$.
The fifth step is to use that homeomorphism to isotope $\gamma''$ to $\delta'$.
The final step is to pull back the isotopy (going first from $\gamma'$ to $\gamma''$, and then from $\gamma''$ to $\delta'$) via the hoeomorphism $f$ to get the desired isotopy from $\gamma$ to $\delta$.