This will not be a complete answer, but it's a different point of view from other stuff posted here. Recall the conventional paramatrization of the torus:
\begin{align} & R > r > 0 \\[6pt] x & = (R + r \cos v) \cos u \\[4pt] y & = (R + r \cos v) \sin u \\[4pt] z & = r \sin v \end{align}
As a mapping from the flat torus $\mathbb{R}^2/(2\pi\mathbb{Z})^2)$ to the embedded torus, this is not conformal. I was suggesting replacing it by
\begin{align} & R > r > 0 \\[6pt] x & = (R + r k(v)) \cos u \\[4pt] y & = (R + r \ell(v)) \sin u \\[4pt] z & = r \sin v \end{align}
where $v\mapsto(k(v),\ell(v))$ is some other parametrization of the unit circle.
Let's call $u$ the longitude and $v$ the latitude. Then the meridians of longitude are closer together on the inside of the embedded torus than on the outside, but the parallels of latitude are still just as far apart for equal values of $\Delta v$. That tells us the mapping is not conformal. So the idea would be to make $(x,y,z)$ move only as fast with respect to $v$ as it moves with respect to $u$. That rate would be proportional to the distance from the center of the embedded torus.
So look at the circle \begin{align} x & = R + r \cos v \\ z & = r \sin v \end{align} i.e. $(x-R)^2 + z^2 = r^2$. Draw any line through $(x,z)=(0,0)$ that intersects that circle twice: once on the "inside", i.e. closer to $(x,z)=(0,0)$ and thus closer to the center of the torus, and once on the "outside". As that line moves, so that both of those two points of intersection move, what is the ratio of their two rates of motion along the arc? Here's an easy result I proved: That ratio of rates of motion is the same as the ratio of their $x$-coordinates! I rashly conjectured this based only on the fact that it's obviously true in the case where the aforementioned line through the origin happens to be the $x$-axis in the $xz$-plane. There didn't seem to be any obvious geometric reason why it should happen, and yet one can verify that it does.
What we would like, then, is that points $v$ and corresponding points $\pi-v$ would be mapped by $ v\mapsto \begin{cases} x = R + r k(v) \\ z = r \ell(v) \end{cases} $ to two points on the circle such that if you draw the line through them it goes straight through the origin in the $xz$-plane.
A corollary is that the point $v=\pi/2$ would get mapped to the point on the circle at which the tangent line goes through the origin.
Although this tells us exactly the mapping $(k(v),\ell(v))\mapsto(k(\pi-v),\ell(\pi-v))$, it stops short of identifying $v\mapsto(k(v),\ell(v))$, except at four special points ($(\pi/2)\mathbb{Z}\mod 2\pi\mathbb{Z}$). Hence this is not a complete answer.