Is there a LLipschitz homeomorphism of the Elipse $x^2/4+y^2=1$ onto the unit circle $x^2+y^2=1$ such that $L<1$?

$\begingroup$ Of course, because the ellipse in question has length greater than that of the circle, so there is a Riemannian homothety (with $L < 1$). $\endgroup$ – Daniele Zuddas May 18 '13 at 22:19

$\begingroup$ The OP must want it to be Lipschitz with respect to the metric induced from the Euclidean plane, not the intrinsic Riemannian metric on the ellipse and circle. It still feels like the answer should still be yes but it's not as immediate. $\endgroup$ – Noam D. Elkies May 19 '13 at 1:17

$\begingroup$ The metric from the Euclidean plane is assumed and Mixon has the answer. $\endgroup$ – djoke May 19 '13 at 7:05
Assuming you mean Lipschitz with respect to the plane's Euclidean metric, as suggested by Noam D. Elkies, then no such homeomorphism exists.
The first thing to worry about is where $(0,\pm1)$ are mapped. Note that they have distance $2$, so a mapping $f$ with $L<1$ will not send these to opposite points on the unit circle. Put $A=f(0,1)$ and $B=f(0,1)$. Then $A$ and $B$ lie in a common semicircle.
Next, we consider $A'$ and $B'$, the points on the unit circle which are opposite $A$ and $B$, respectively. Specifically, parameterize the portion of the circle from $A$ to $B'$ as $C(t)$ and from $A'$ to $B$ as $C'(t)$ such that $C(t)$ and $C'(t)$ are opposite points on the circle for every $t$. Then for each $t$, you need $\f^{1}(C(t))f^{1}(C'(t))\>2$. Note that by the homeomorphism, $f^{1}(A')$ and $f^{1}(B')$ lie on the same side of the $y$axis. Then by the intermediate value theorem, there is a $t$ such that $f^{1}(C(t))$ and $f^{1}(C'(t))$ have the same $x$coordinate. But that implies $\f^{1}(C(t))f^{1}(C'(t))\\leq2$.