Example of an invariant metric on a nilpotent group which is not asymptotically geodesic Let $X$ be a metric space. We say that $X$ is asymptotically geodesic if for all $\epsilon > 0$, there exists $R > 0$ such that, for all $x,y \in X$, there exists some finite sequence of points $p_0=x,p_1,p_2,...,p_N = y \in X$ such that each $d(p_{i-1},p_i) \le R$ and
$$ \sum_{i=1}^N d(p_{i-1},p_i) \le (1 + \epsilon)d(x,y). $$
I came across this definition while studying finitely generated nilpotent groups and their scaling limits. Pansu proved in "Croissance des boules et des geodesiques fermees dans les nilvarietes" that, if a finitely generated (virtually) nilpotent group $\Gamma$ is endowed with a left-invariant metric $d$ such that $(\Gamma,d)$ is asymptotically geodesic, then the sequence of metric spaces $(\Gamma,\frac{1}{n}d)$ has a Gromov-Hausdorff limit (which he describes).
However, I've been having a lot of difficulties coming up with examples of metrics on nilpotent groups which aren't asymptotically geodesic, particularly under the extra assumption that the metric in question is bi-Lipschitz to a word metric on $\Gamma$.
I believe that it follows pretty quickly from Fekete's lemma that any invariant metric on $\mathbb{Z}$ which is bi-Lipschitz to the standard metric is asymptotically geodesic, and, unless I made a mistake, the same is true for $\mathbb{Z}^d$ by a more involved but fairly similar argument. But the property of being asymptotically geodesic is certainly not preserved by bi-Lipschitz maps a priori, and it seems implicit in the literature that there should be some metrics bi-Lipschitz to word metrics which are not asymptotically geodesic. (For instance, in Benjamini and Tessera, "First passage percolation in nilpotent Cayley graphs and beyond", one of the key ingredients in a proof is showing that a certain invariant metric is asymptotically geodesic, even though that metric is assumed to be bi-Lipschitz to a word metric).
So my question is:
(1) What is an example of a left-invariant metric on a finitely generated nilpotent group which is bi-Lipschitz to a word metric but not asymptotically geodesic?
As it happens, I actually have had difficulty constructing such a metric even if I drop the assumption that it is bi-Lipschitz to a word metric, so I would also be interested in the weaker question:
(2) What is an example of a left-invariant metric on a finitely generated nilpotent group which is not asymptotically geodesic?
And even though I'm interested primarily in the case of nilpotent groups, I'm also generally trying to understand this property, so an answer to the even weaker question might help too:
(3) What is an example of a left-invariant metric on a (finitely generated?) discrete group which is not asymptotically geodesic?
 A: This answer seems to essentially be what @YCor was going for in his comments, but I'll give very explicit examples, since at the level he described the answer, there were some details that needed to be checked that weren't immediately clear to me. (I hope this is an appropriate situation to answer my own question.)
First, without any requirement of being bi-Lipschitz to a word metric, @YCor's first comment suggests a very easy example on $\mathbb{Z}$, that is,
$d(m,n) = \sqrt{|n-m|}$. It's helpful to remember that the square root of any metric is a metric, and taking the square root is a nice way to ruin asymptotic geodesicity. Anyone reading this should work out why asymptotic geodesicity fails in this example in order to understand the next one.
Now, for an example of a metric which is bi-Lipschitz to a word metric but not asymptotically geodesic (suggested by @YCor's second comment).
Take $\Gamma$ to be the Heisenberg group
$$ \Gamma := \langle X, Y, Z | [X,Y]=Z, [X,Z]=[Y,Z]=1\rangle.$$
It's not too hard to show that each element of $\Gamma$ can be written uniquely
as $X^k Y^l Z^m$, and so we have a bijection $X^k Y^l Z^m \leftrightarrow (k,l,m)$
between $\Gamma$ and $\mathbb{Z}^3$.
It's also a standard exercise to show that, for some constants $0<c<C<\infty$,
$$ c \max(|k|,|l|,\sqrt{|m|}) \le |(k,l,m)| \le C \max(|k|,|l|,\sqrt{|m|}), $$
where $|(k,l,m)|$ is the length of the shortest word in $X$ and $Y$ which is equal to $X^k Y^l Z^m$. (If you haven't done this before, the key observation is that $|Z^{m^2}| = |[X^m,Y^m]| = O(m)$. It's also helpful in the analysis to have the following interpretation of the geometry of the Heisenberg group: a word in $X,Y$ represents a path in the Cayley graph of $\Gamma$ with respect to the generators $X$ and $Y$, and we can look at the projection of this path onto the Cayley graph of the abelianization
$\Gamma^{ab} \cong 
\langle \bar{X}, \bar{Y} | [\bar{X},\bar{Y}]=1 \rangle \cong \mathbb{Z}^2$.
If the word is equal to $Z^m$, then the projected path in $\mathbb{Z}^2$ is closed and $m$ is equal to the (signed) area enclosed by that path; for example $[X^m,Y^m]=Z^{m^2}$ draws out a square in $\mathbb{Z}^2$ with area $m^2$.)
One then has to check that (1) $\max(|k|,|l|,\sqrt{|m|})$ is in fact subadditive on $\Gamma$, and hence induces an invariant metric $d(x,y)=|x^{-1}y|$ on $\Gamma$ which is bi-Lipschitz to the word metric, and (2) the induced metric $d$ is not asymptotically geodesic. This should be true, but if one is concerned that they might have made some errors in their analysis, note also that for any $\epsilon > 0$,
the function $\max(|k|,|l|,\epsilon\sqrt{|m|})$ is also bi-Lipschitz, and the smaller $\epsilon$ is, the easier it is to confirm that this function is subadditive and that it is not asymptotically geodesic. Here the heuristic to see that this metric is not asymptotically geodesic is the following: one cannot get an approximate geodesic from the identity to $Z^M$ by traveling along powers of $Z$ for the same reason that the square root of the standard metric on $\mathbb{Z}$ is not asymptotically geodesic. The paths which should approximate geodesics should come from words in $X$ and $Y$, but the length of such paths will be roughly some constant (independent of $\epsilon$) times $\sqrt{M}$, while the prescribed distance from the identity to $Z^M$ is
$\epsilon \sqrt{M}$.
