Isometry from Manhattan plane to Euclidean plane?

Does there exist an isometry from a Manhattan plane $A$ to a Euclidean plane $B$? I.e. a function $\varphi:A \to B$ that suffices $\|\varphi(a)\|_B = \|a\|_A$ for all $a \in A$, where $\| \cdot \|_A$ is the Manhattan norm and $\| \cdot \|_B$ the Euclidean norm.

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Hint: In the Euclidean plane, for a pair of points $x,y$ with distance $2$, how many points are there at distance $1$ from both $x$ and $y$? What about the Manhattan plane? (Look at $x = (0,0)$ and $y = (1,1)$.)

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I guess not. Does this still hold for the euclidean $\mathbb{R}^3$ instead of $\mathbb{R}^2$? – Chiel ten Brinke Apr 9 '13 at 11:15
@Chiel92: I've modified my answer. If you understand this hint, you also have the solution for higher dimension. – azimut Apr 9 '13 at 11:18
These three points can be mapped with preserved mutual distances, right? I guess I don't get this hint, while I did get the previous hint. – Chiel ten Brinke Apr 9 '13 at 11:21
@Chiel92: Thanks for accepting! I've edited my answer again, in the hope to get at the core or the argument. – azimut Apr 9 '13 at 11:33

I actually had the same question, and the answer is that there is no isometry, but you can find a relation between both norms. It takes to add a second variable to the norm, that is the number $N(a)$ of minimal paths to cover $a$ on the manhattan plane.

$||a||_B = \phi(||a||_A,N(a))$ or $||a||_A = \psi(||a||_B,N(a))$

You can verify easily on a few examples that it works, and if you compute the number of paths, you will find out that as long as you are just after the euclidean norm. If you compute the number of paths, you'll find explicit forms for $\phi$ and $\psi$.

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