Let $B^n$ be the unit open ball in $\mathbb{R}^n$, $p\in \mathbb{R}^n$ and $f\colon \mathbb{R}^n\to B^n$ defined as $f(x)=\frac{1}{1+|x|}x$.
I believe there are constants $C_p>0$ such that $|f(x+p)-f(y+p)|\le C_p|f(x)-f(y)|$ for all $x,y\in \mathbb{R}^n$. I tried things such as $C_p=1+|p|$ or $C_p=(1+|p|)^2$ (these seem good candidates) but I'm having problems with some estimates. Any other constant works, of course.
Edit: Some advance in the problem:
Note first that $f$ is actually an homeomorphism from $R^n$ to $B^n$ so in particular it's injective. So we are to find a constant $C_p$ such that $|f(x+p)-f(y+p)|\le C_p|f(x)-f(y)|$ for all $x,y\in \mathbb{R}^n$. We may assume that $x\neq y$ and so the last inequality becomes \begin{equation} \frac{(1+|x|)(1+|y|)}{(1+|x+p|)(1+|y+p|)}\frac{|(1+|y+p|)(x+p)-(1+|x+p|)(y+p)|}{|(1+|y|)x-(1+|x|)y|}\le C_p \end{equation}
But note that $(1+|x|)/(1+|x+p|)\le 1+|p|$ and similar for $y$ so we have only to find a constant $Q_p$ such that \begin{equation*} \frac{|(1+|y+p|)(x+p)-(1+|x+p|)(y+p)|}{|(1+|y|)x-(1+|x|)y|}\le Q_p \end{equation*}
Edit 2: More advances in the problem
Let's assume that $p\neq 0$. I think that $Q_p=1+2|p|$ and, in fact, that $1+2|p|$ can't be improved. A straightforward calculation shows that \begin{align*} (1+|y+p|)(x+p)-(1+|x+p|)(y+p)=&|y+p|(x+p)-|x+p|(y+p)+|x|y-|y|x\\ &+(1+|y|)x-(1+|x|)y \end{align*} so that \begin{equation*} \frac{|(1+|y+p|)(x+p)-(1+|x+p|)(y+p)|}{|(1+|y|)x-(1+|x|)y|}\le\frac{||y+p|(x+p)-|x+p|(y+p)+|x|y-|y|x|}{|(1+|y|)x-(1+|x|)y||}+1 \end{equation*} I claim that \begin{equation*} \frac{||y+p|(x+p)-|x+p|(y+p)+|x|y-|y|x|}{|(1+|y|)x-(1+|x|)y|}<2|p| \end{equation*} and that this bound can't be improved because taking $y=0$ and $x=-\lambda p$ and letting $\lambda\to \infty$ we see that the left hand side of the above equation tends to $2|p|$.
However I can't prove the last inequality. Some numerical tests with Mathematica tell me that it's true though.