A metric on $\mathbb{R}^n$ such that $d(\lambda x, \lambda y)=|\lambda| d(x,y)$ which is not induced by a norm

Question

Let $V=\mathbb{R}^n$.
Let $d:V \times V\rightarrow \mathbb{R}$ a metric on $\mathbb{R}^n$.
Assume that for any $x,y\in V$ and $\lambda \in \mathbb{R}$, we have $d(\lambda x, \lambda y) = |\lambda|d(x,y)$.
Is $d$ necessarily induced by a norm?

Motivation: I've been thinking of $\pi$ and thought about why the ratio between a circles's circumference and its radius is constant. The proof is easy and is applicable to any norm. I think the "positive homogeneity" condition I posed on the metric above is enough for this ratio to be constant.

Answer 1

The answer is no. You need translational invariance as well; then it's a pretty well-known theorem (see e.g. here).

As a counterexample when leaving out the translational invariance, consider:

$$d: \Bbb R^n \times \Bbb R^n \to \Bbb R_{\ge 0}: d (x,y)=\begin{cases} \|x\|+\|y\| & \text{if $x \ne y$}\\ 0 & \text{otherwise.} \end{cases}$$

This metric is sometimes referred to as the "metric of the French railway system", although there are similar metrics with the same name (cf. the comments).