Assume that $\gamma$ is a convex curve which is symmetric with respect to two orthogonal lines (the ellipse is such a curve).
I want to know if there exists a $(l,L)$-bi-Lipschitz mapping of the unit circle onto $\gamma$ such that $L=\mathbf{diam}(\gamma)/2$ and $l=\mathbf{dist}(\gamma,0)$, where $0$ is the center of $\gamma$.
Here is a weaker version of the above question: Does there exist a $L$-Lipschitz homeomorphism of the unit circle onto $\gamma$ such that $L=\mathbf{diam}(\gamma)/2$?
Consider the convex hull $L_\varepsilon$ of the points $(\pm 1,0)$ and the $\varepsilon$-disc centered at the origin for small $\varepsilon>0$.
If there is a $(\varepsilon,1)$-bi-Lipschitz map from the unit disc onto $L_\varepsilon$ then all the angles of curve bounding $L_\varepsilon$ has to be at least $\pi{\cdot}\varepsilon$ (see the sketch below).
On the other hand, $L_\varepsilon$ has two corners with angles about $2{\cdot} \varepsilon$.
The sketch. Assume $f$ maps the half-plane $H$ to the angle $A$ and has Lipschitz constants $\varepsilon$ and $1$, yet assume that $f(0)=0$ and $0$ is the tip of $A$.
Essentially we need to show that angle measure of $A$ is at least $\varepsilon{\cdot}\pi$. Note that if $B_r(x)$ lies in $H$ then $B_{\varepsilon\cdot r}(f(x))$ lies in $A$. It follows that the image of an $\alpha$-angle in $H$ with tip at $0$ intersects unit circle along an arc of length $\varepsilon{\cdot}\alpha+o(\alpha)$. Hence the result follows.
For the Lipschitz-only part, the answer is yes. More generally, if $\alpha$ and $\beta$ are any two convex curves and $\beta$ fits inside an $L$-homothetic copy of $\alpha$, then there exist an $L$-Lipschitz map $f$ from $\alpha$ onto $\beta$.
Indeed, by rescaling we may assume that $\beta$ is inside $\alpha$, and $L=1$. In this case, let $f$ be the nearest-point projection to $\beta$, i.e. $f(x)$ is the point on $\beta$ nearest to $x$. It is easy to see that this map is well-defined in the outer region of $\beta$ and does not increase distances.
