I've come to a problem I'm not quite sure how to tackle, but I realise it's quite simple in its core.
Let's assume $(M,d_2)$, where M is a set of vertices in $R^2$ space, and d being Euclidean metric. Now let's assume another metric space $(M,d_1)$, M being the same set of vertices in 2D space and $d_1$ being Manhattan (Taxicab) metric.
The question is, are these two metric systems isomorphic?
PS: I'm not quite sure if this question is correctly formalized. Just in case I'm adding my personal problem to minimize any confusion.
I'm building a simple code to compare vertex distances in 2D space, but Euclidean distance is computationally too difficult for my needs, because I only need comparation. Thus, I'd like to use Manhattan metric system for its simplicity, but I don't know if I can't end up with a solution that's false in Euclidean metrics.
The 2 metrics do not have the property you want (if nothing more is known about the potential positions of the vertices). For example, let's compare the distances of the points $P_1=(3,4)$ and $P_2=(0,6)$ to the origin $O=(0,0)$.
For the Manhattan metric $d_1$ we get $d_1(O,P_1) = 3+4 = 7$, and $d_1(O,P_2) = 6$.
For the Euclidean metric $d_2$ however we get $d_2(O,P_1) = \sqrt{3^2+4^2} = 5$, and $d_2(O,P_2) = 6$.
This means that the distance from $O$ to $P_1$ is smaller than the distance from $O$ to $P_2$ for the Euclidean metric but greater for the Manhattan metric.
To get back to your stated motivation, usually the numerical cost of the Euclidean distance is in calculating the square root. If you compare the square of the Euclidean distance $$d^2_2((x_1,y_1),(x_2,y_2))=(x_1-x_2)^2 + (y_1-y_2)^2$$ this preserves the ordering of distances with respect to the Euclidean distance (because the square function is monotonically increasing for positive arguments).