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 $L<1$?

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 $\f^{1}(C(t))f^{1}(C'(t))\\leq2$. 

