Function $T:X \rightarrow X$ with least Lipschitz constant, for which $\sup_{x \in X}d(T(x),x) \le r$. Let $X=(X,d)$ be a metric space and $r \ge 0$.

Question 1. What is a good upper-bound (the smaller the better) for $\inf_T Lip(T)$ where $Lip(T) := \underset{x,x' \in X,\;x \ne x'}{\sup}\; \frac{d(T(x),T(x'))}{d(x,x')}$, and the infimum is taken over all functions $T:X \rightarrow X$ such that $\underset{x \in X}{\sup}\;d(T(x),x) \le r$ ?
Question 2. Same question when $X$ is a normed vector space, e.g $\mathbb R^n$ equipped with an $\ell_p$-norm, especially the cases $p=1,2,\infty$.

N.B.: Of course, these bounds would depend on the parameter $r$. Also note that the an upper bound of $1$ is trivial and uninformative (gotten by taken $T =I_X$, the identity function on $X$).
 A: If $r\geq\sup_{x,x'\in X} d(x,x')$, then the infimum is equal to $0$ (as we can simply take a constant function. 
On the other hand, if $\sup_{x,x'\in X} d(x,x')=\infty$, then your infimum is equal to 1. This we see from
$$ \frac{d(T(x),T(x'))}{d(x,x')} \geq \frac{d(x,x')-d(T(x),x) -d(x', T(x')) }{d(x,x')} \geq 1 - \frac{2r}{d(x,x')} $$
and letting $d(x,x') \rightarrow \infty$ yields the desired result (when combined with your trivial upper bound). In particular this gives you the result for any (non-trivial) normed space.
My guess is that for general metric spaces it becomes extremely nasty (just play around with some metric spaces consisting of finitely many points). If you consider finite metric spaces and take $r< \min_{x,x'\in X \ : \ x\neq x'} d(x,x')$, then your infimum will be equal to $1$. On the other hand if it is bigger, then it will depend on the relative position of those points to each other. 
Added: The computation above shows that the infimum above is always bounded from below by $\max\{0, 1-2r/diam(X)\}$. So that would be best we can get and sometimes we get it (for example for intervals with the Euclidean metric or for all normed vector spaces). However, considering finite metric spaces shows that this lower bound is not always attained.
Added: Another example that in general we cannot hope to get a nice formula. Consider the circle $S^1 \subset \mathbb{C}$. We define the metric via the length of the minimal connecting curve. Now pick $r<\pi/2$. We show that the infimum over the lipschitz constants is equal to $1$.
Assume by contradiction that there exists a contraction $T$ on $S^1$ satisfying the condition with $r<\pi/2$.
Let us define 
$$A_1:=\{ e^{2\pi x} \in \mathbb{C} \ : \ x\in (0;1/2)\}, \quad A_2:=\{ e^{2\pi x} \in \mathbb{C} \ : \ x\in (1/2;1)\}$$
$$ A_3:=\{ e^{2\pi x} \in \mathbb{C} \ : \ x\in (-1;-1/2)\}, \quad  A_4:=\{ e^{2\pi x} \in \mathbb{C} \ : \ x\in (-1/2;0)\}  $$
Wlog we may assume that $T(1)=1$ (otherwise precompose with a rotation). In addition we may assume wlog $T(-1)\in A_1 \cup A_2$. But from $r<\pi/2$ we get $T(-1)\in A_2$.
As $1$ is a fixed point and $T$ is a contraction, we get $T(-i)\in A_1 \cup A_4$. However, as $r<\pi/2$, we must have $T(-i)\in A_4$.
As $T(-1)\in A_2$ and $T(-i)\in A_4$ we get $d(T(-1), T(-i)) > \pi/2$ and therefore
$$ \frac{d(T(-1), T(-i))}{d(-1, -i)} > \frac{\pi/2}{\pi/2} =1. $$
This is a contradiction as $T$ is a contraction.
Added: On the other hand, if we consider the metric space $(-1;1)$ with the Euclidean metric, then we have the map $T: (-1;1) \rightarrow (-1;1), \ T(x) = \left(1-r \right)x$. We have $\vert T(x) - x\vert = r \vert x \vert < r$ and $Lip(T)=1-r = 1-\frac{2r}{diam((-1;1))}$.
