Finding a metric to make a certain curve a circle Given a  simple closed, regular $C^\infty$ curve $\phi$ in $U\subset\mathbb{R}^n$ naturally parametrized (by it's arc length), is there any way to
obtain a Riemaniann manifold $(S,g)$  of dimension 2 without boundary (isometrically embedded in $\mathbb{R}^m$ equipped with the standard metric) such that there is a geodesic circle in this surface that is equal to the curve (meaning that it is mapped by the embedding to $\phi$)?
Two examples:
$1)\gamma(t)=\begin{pmatrix} \left(\frac{\sin(20\pi t)}{10}+1\right)\sin(2\pi t)\\ \left(\frac{\sin(20\pi t)}{10}+1\right) \cos(2\pi t)\\ \sin(2\pi t)\end{pmatrix}\\
2)\gamma(t)=\begin{pmatrix} \left(\frac{\sin(20\pi t)}{10}+1\right)\sin(2\pi t)\\ \left(\frac{\sin(20\pi t)}{10}+1\right) \cos(2\pi t)\end{pmatrix}$
One possibility, considering $\gamma \in \mathbb{R}^2$, is to use the Riemann smooth mapping theorem in such a way to obtain a complex diffeomorphism $\phi$ between $\gamma\bigcup \text{Int}(\gamma)$ and the closed unitary disk $D$. In this way, we might define the metric tensor on $S=\phi^{-1}(D)$ as the pullback of the euclidean metric tensor restricted to the unitary disk, but that leaves us with a manifold with boundary. We may try to extend it, but such a subject is quite technical, and I would not know how to proceed. Even if this idea was succesful This method would work only in $\mathbb{R}^2$, leaving open the question for $n>2$.
The questions are thus:
1) Is my idea efficient to solve the problem in $\mathbb{R}^2$? If so, how to remove the boundary?
2)How to attack the problem if $\gamma \subset \mathbb{R}^n$ with $n>2$ (as an example, see the first example)?
 A: This is not a complete answer; just a few potentially useful thoughts.
Throughout, I'll use $\mathbb{R}^n$ for the manifold $(\mathbb{R}^n,g)$ to denote an arbitrary Riemannian metric on $\mathbb{R}^n$ and $\mathbb{E}^n$ to denote $\mathbb{R}^n$ with the standard metric.
I'll interpret the question (with some further restrictions) as follows.

Def. Given a closed curve $\gamma:S_1\to\mathbb{E}^n$, we say $\gamma$ is an extendable disc boundary there exists an open 2-ball $B$ containing a concentric closed 2-disc $D$ and an embedding $\iota:B\to\mathbb{E}^n$ such that $\iota(\partial D)=\gamma(S_1)$. Further, it is a extendable geodesic disc boundary if $B$ and $D$ can be made geodesic balls/discs with the induced metric.

We can it seems forget about the open 2-ball entirely, and just look at the closed disc.

Def. Given a closed curve $\gamma:S_1\to\mathbb{E}^n$, we say $\gamma$ is a disc boundary if there exists a closed 2-disc $D$ and an embedding $\iota:D\to\mathbb{E}^n$ such that $\iota(\partial D)=\gamma(S_1)$. If is a geodesic disc boundary if $B$ can be made a geodesic disc in the induced metric.

It turns out these two conditions are equivalent.

Prop. All (geodesic) disc boundaries are extendable.
Proof (sketch). We need only show that a closed disc boundary has an embedded extension. If the disc is isometrically embedded, equipping the extension with the pullback metric satisfies the extendable geodesic disc boundary condition. Let $D$ be an embedded disc parameterized by coordinates $x,y$ with $x^2+y^2\le r^2$. Schematically, I think one can construct an extension as follows:
  
  
*
  
*Construct a closed, embedded tubular neighborhood $T\supset D$ and extend the normal coordinates functions $\nu^2,...,\nu^n$ to all of $\mathbb{E}^n$
  
*Show there is an open neighborhood $\mathcal{O}_T\supset T$ such that the level set $\nu^i=0$ is an embedded submanifold $\hat{D}$ through level set theorem and rank arguments.
  
*Extending the coordinate functions $x,y$ to $\hat{D}$. We know that the function $(x,y):\hat{D}\to\mathbb{R}^2$ is invertible when restricted to $D$, so there is an open set in $\hat{D}$ on which the function is invertible (see this answer for an idea of how to show this). This open set contains an open neighborhood containing $D$ diffeomorphic to an open disc.
  

Still, there are topological, differential, and geometric obstructions to this construction. The general requirement is that the curve be an embedding of $S_1$, i.e. has nonvanishing velocity, since the boundary of a disk is precisely such an embedding. For additional restrictions, we can proceed by dimension.
In $n=0,1$, the disc embedding is trivially impossible by rank considerations.
In $n=2$, given any closed, embedded curve $\gamma(S_1)$, is automatically the boundary of exactly one closed disc, so it suffices to check if this disc with the pullback metric is a geodesic disc. If $\gamma(S_1)$ is already geodesic circle, this is automatically true, since circles in $\mathbb{E}^2$ are geodesically convex. If a curve is not a circle, it is not an isometrically embedded disc: we can choose a point $c$ encircled by the curve, find a point $x$ which minimizes the distance function, and find some other point $y$ with $d(c,x)<d(c,y)$. Restricting to the disc encircled by $\gamma(S_1)$ preserves this inequality.
In $n=3$, the curve cannot be knotted. There is a result from topology that a knotted curve cannot be the boundary of a contractible embedded surface (in particular, a disc).
For unknotted curves $n\ge 3$ (which includes all curves in $n\ge 4$), we can always find an embedded disc $D$, but this disc need not be geodesic.
To find a geodesic disc, we will need to "deform" $D$ to have the right induced metric. A few possibilities come to mind for how to do this. The most straightforward try is probably to work in a tubular neighborhood with coordinates $r,\theta,\nu^1,\dots,\nu^n$ such that $D=\{r\le 1,\nu^i=0\}$ and deform it to a "graph" $D'=\{r\le 1,\nu^i=\nu^i(r,\theta)\}$. From there, one can write the metric as a function $\nu^i$, find a family of deformations $\nu^i(r,\theta)$, and try to change the metric on $D'$ to satisfy sufficient conditions for a geodesic circle.
