Is there an Isometry between metrics? Is there an isometry between the taxi-cab metric and the max metric? How about the euclidean metric to the taxi-cab metric?
 A: The taxi-cab metric is isometric to the max metric only for dimensions $1$ and $2$, and for higher dimensions they are not isometric because their unit balls have different number of extreme points (isometries of normed spaces carry extreme points to extreme points of the corresponding unit balls). The Euclidean metric is not isometric to none of the above in any dimension, again, by an extreme point argument: Every point on the boundary of a Euclidean ball is extreme, whereas the other two unit-balls have only a finite number of extreme points in any dimension.
A: I would not word things this way. One metric space is $\mathbb R^2$ with the $L^\infty$ metric, that is the maximum absolute value of the two coordinates. The other is  $\mathbb R^2$ with the $L^1$ metric, sum of the absolute values. There is an isometry between them, simply rotate by $45^\circ$ and expand (dilate) by a factor of $\sqrt 2.$ In the opposite direction, shrink by the same factor. The mapping is very nice and linear, it just does not preserve the $L^2$ metric. 
Anyway, an isometry is a mapping $f$ between metric spaces, with
$$ d_2(f(u), f(v) ) = d_1(u,v) $$
