OK, here is why $Lip1$ is impossible. Suppose we have such choice. Then consider all $K$ whose minimal containing box is a unit square. Let $[x(K),x(K)+1]\times[y(K),y(K)+1]$ be the minimal containing box of $K$. Let $e=(1,0)$. Note that if $L$ is another such domain, then for fixed $t$ and $T\to+\infty$, we have $d_H(te+K,Te+L)=(T-t)+x(L)-x(K)+o(1)$ (the fixed vertical shift becomes of no importance). Thus, we conclude that $p(te+K).x-x(te+K)=w$, then $p(Te+L).x-x(Te+L)\le w+o(1)$ as $T\to\infty$, whence the difference $p(Te+K).x-x(Te+K)$ has a (usual) limit $w_x$ independent of $K$ as $T\to+\infty$. Now if we define $$ p_T(K)=p(Te+K)-Te $$
and put $p_1(K)=\lim_T p_T(K)$ where the limit is taken with respect to some ultrafilter containing the usual filter of rays going to $+\infty$, we shall get a choice whose $x$-coordinate is $x(K)+w_x$ for every $K$ in our class and that choice still is $Lip1$. Now we can run the same construction in the $y$-direction and conclude that we can make a $Lip1$ choice $p_2$ such that $p(K).y=y(K)+w_y$ as well. But then for this class of domains, choosing the box center works as well. However if we take $K=conv((-1,0),(0,-1),(0,0))$ and $L=-K$ and observe that $d_H(K,L)=1$ but the box shift is of length $\sqrt2$.

Remarks:

1. The ultrafilter is not really needed but it makes the proof very transparent, so I decided to use it. Those who dislike AC are welcome to recast it as an argument with classical limits only (or, better, without limits at all using the pigeonhole principle instead) as an exercise.

2. While this argument shows that the minimal possible Lipschitz constant is strictly greater than $1$ (if we could get it arbitrarily close to $1$, we could get $1$ as well by taking the limit along, erm, some ultrafilter on positive integers containing the standard filter for the usual limit), it does not provide any explicit lower bound. So finding a reasonable lower bound $1+\delta$ for the Lipschits constant remains an open problem (some quantification is actually possible even using just finitely many shifts of the last two triangles, but I would rather not write down the ridiculously small $\delta$ it yields).

3. I hope that is all correct, but check everything carefully because it is quite late here now and I'm not at my best today :-)

OK, here is why $Lip1$ is impossible. Suppose we have such choice. Then consider all $K$ whose minimal containing box is a unit square. Let $[x(K),x(K)+1]\times[y(K),y(K)+1]$ be the minimal containing box of $K$. Let $e=(1,0)$. Note that if $L$ is another such domain, then for fixed $t$ and $T\to+\infty$, we have $d_H(te+K,Te+L)=(T-t)+x(L)-x(K)+o(1)$ (the fixed vertical shift becomes of no importance). Thus, we conclude that if $p(te+K).x-x(te+K)=w$, then $p(Te+L).x-x(Te+L)\le w+o(1)$ as $T\to\infty$, whence the difference $p(Te+K).x-x(Te+K)$ has a (usual) limit $w_x$ independent of $K$ as $T\to+\infty$. Now if we define $$ p_T(K)=p(Te+K)-Te $$
and put $p_1(K)=\lim_T p_T(K)$ where the limit is taken with respect to some ultrafilter containing the usual filter of rays going to $+\infty$, we shall get a choice whose $x$-coordinate is $x(K)+w_x$ for every $K$ in our class and that choice still is $Lip1$. Now we can run the same construction in the $y$-direction and conclude that we can make a $Lip1$ choice $p_2$ such that $p(K).y=y(K)+w_y$ as well. But then for this class of domains, choosing the box center works as well. However if we take $K=conv((-1,0),(0,-1),(0,0))$ and $L=-K$ and observe that $d_H(K,L)=1$ but the box shift is of length $\sqrt2$.

Remarks:

1. The ultrafilter is not really needed but it makes the proof very transparent, so I decided to use it. Those who dislike AC are welcome to recast it as an argument with classical limits only (or, better, without limits at all using the pigeonhole principle instead) as an exercise.

2. While this argument shows that the minimal possible Lipschitz constant is strictly greater than $1$ (if we could get it arbitrarily close to $1$, we could get $1$ as well by taking the limit along, erm, some ultrafilter on positive integers containing the standard filter for the usual limit), it does not provide any explicit lower bound. So finding a reasonable lower bound $1+\delta$ for the Lipschits constant remains an open problem (some quantification is actually possible even using just finitely many shifts of the last two triangles, but I would rather not write down the ridiculously small $\delta$ it yields).

3. I hope that is all correct, but check everything carefully because it is quite late here now and I'm not at my best today :-)

OK, here is why $Lip1$ is impossible. Suppose we have such choice. Then consider all $K$ whose minimal containing box is a unit square. Let $[x(K),x(K)+1]\times[y(K),y(K)+1]$ be the minimal containing box of $K$. Let $e=(1,0)$. Note that if $L$ is another such domain, then for fixed $t$ and $T\to+\infty$, we have $d_H(te+K,Te+L)=(T-t)+x(L)-x(K)+o(1)$ (the fixed vertical shift becomes of no importance). Thus, we conclude that $p(te+K).x-x(te+K)=w$, then $p(Te+L).x-x(Te+L)\le w+o(1)$ as $T\to\infty$, whence the difference $p(Te+K).x-x(Te+K)$ has a (usual) limit $w_x$ independent of $K$ as $T\to+\infty$. Now if we define $$ p_T(K)=p(Te+K)-Te $$
and put $p_1(K)=\lim_T p_T(K)$ where the limit is taken with respect to some ultrafilter containing the usual filter of rays going to $+\infty$, we shall get a choice whose $x$-coordinate is $x(K)+w_x$ for every $K$ in our class and that choice still is $Lip1$. Now we can run the same construction in the $y$-direction and conclude that we can make a $Lip1$ choice $p_2$ such that $p(K).y=y(K)+w_y$ as well. But then for this class of domains, choosing the box center works as well. However if we take $K=conv((-1,0),(0,-1),(0,0))$ and $L=-K$ and observe that $d_H(K,L)=1$ but the box shift is of length $\sqrt2$.

Remarks:

1. The ultrafilter is not really needed but it makes the construction very transparent, so I decided to use it. Those who dislike AC are welcome to recast it as an argument with classical limits only (or, better, without limits at all using the pigeonhole principle instead) as an exercise.

2. While this argument shows that the minimal possible Lipschitz constant is strictly greater than $1$ (if we could get it arbitrarily close to $1$, we could get $1$ as well by taking the limit along, erm, some ultrafilter on positive integers containing the standard filter for the usual limit), it does not provide any explicit lower bound. So finding a reasonable lower bound $1+\delta$ for the Lipschits constantremains an open problem (some quantification is actually possible even using just finitely many shifts of the last two triangles, but I would rather not write down the ridiculously small $\delta$ it yields).

3. I hope that is all correct, but check everything carefully because it is quite late here now and I'm not at my best today :-)

OK, here is why $Lip1$ is impossible. Suppose we have such choice. Then consider all $K$ whose minimal containing box is a unit square. Let $[x(K),x(K)+1]\times[y(K),y(K)+1]$ be the minimal containing box of $K$. Let $e=(1,0)$. Note that if $L$ is another such domain, then for fixed $t$ and $T\to+\infty$, we have $d_H(te+K,Te+L)=(T-t)+x(L)-x(K)+o(1)$ (the fixed vertical shift becomes of no importance). Thus, we conclude that $p(te+K).x-x(te+K)=w$, then $p(Te+L).x-x(Te+L)\le w+o(1)$ as $T\to\infty$, whence the difference $p(Te+K).x-x(Te+K)$ has a (usual) limit $w_x$ independent of $K$ as $T\to+\infty$. Now if we define $$ p_T(K)=p(Te+K)-Te $$
and put $p_1(K)=\lim_T p_T(K)$ where the limit is taken with respect to some ultrafilter containing the usual filter of rays going to $+\infty$, we shall get a choice whose $x$-coordinate is $x(K)+w_x$ for every $K$ in our class and that choice still is $Lip1$. Now we can run the same construction in the $y$-direction and conclude that we can make a $Lip1$ choice $p_2$ such that $p(K).y=y(K)+w_y$ as well. But then for this class of domains, choosing the box center works as well. However if we take $K=conv((-1,0),(0,-1),(0,0))$ and $L=-K$ and observe that $d_H(K,L)=1$ but the box shift is of length $\sqrt2$.

Remarks:

1. The ultrafilter is not really needed but it makes the proof very transparent, so I decided to use it. Those who dislike AC are welcome to recast it as an argument with classical limits only (or, better, without limits at all using the pigeonhole principle instead) as an exercise.

2. While this argument shows that the minimal possible Lipschitz constant is strictly greater than $1$ (if we could get it arbitrarily close to $1$, we could get $1$ as well by taking the limit along, erm, some ultrafilter on positive integers containing the standard filter for the usual limit), it does not provide any explicit lower bound. So finding a reasonable lower bound $1+\delta$ for the Lipschits constant remains an open problem (some quantification is actually possible even using just finitely many shifts of the last two triangles, but I would rather not write down the ridiculously small $\delta$ it yields).

3. I hope that is all correct, but check everything carefully because it is quite late here now and I'm not at my best today :-)