The metric cannot be recovered from its Hausdorff measure in general. Now, assume that $(X,d\_X)$ and $(Y, d\_Y)$ are connected compact length spaces and induce $n$-dimensional Hausdorff measures $\mathcal{H}^n\_X$ and $\mathcal{H}^n\_Y$.


Assume there exists a 1-Lipschitz map $f: (X,d\_X,\mathcal{H}^n\_X)\to (Y, d\_Y, \mathcal{H}^n\_Y)$ such that $f\_\*\mathcal{H}^n\_X=\mathcal{H}^n\_Y$, then, whether $f$ is an isometric map?


Edit: Thanks to Moishe's comment, now let $X$ and $Y$ be closed topological $n$-manifolds. In fact, I wish to show that claim is true for closed Riemannian $n$-manifolds under the condition of $f\_\*\mathcal{H}^n\_X(X)=\mathcal{H}^n\_Y(Y)$.



Recovering the length metric from Hausdorff measure
Nan Li proved that it holds for a pair of Alexandrov spaces without boundary;
in particular, it solves the problem for Riemannian manifolds.
See [Lipschitz-Volume rigidity in Alexandrov geometry](https://arxiv.org/abs/1110.5498).


It seems that his argument can be generalized quite a bit, but one cannot expect it to work for topological manifolds with intrinsic metrics.
Indeed, take the standard sphere $X$. Shrink its equator by factor 2; denote the obtained space by $Y$. The induced map $X\to Y$ is measure-preserving and short.


