Recovering the length metric from Hausdorff measure 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 $f_*\mathcal{H}^n_X(X)=\mathcal{H}^n_Y(Y)$.
 A: The easiest argument I know (which works for path-metrics on topological manifolds $X$ and in even greater generality) is to consider the induced map $f^2: X^2\to Y^2$. This map also preserves the product measure. Set
$$
\Delta_r(X)=\{(x,y)\in X^2: d(x,y)\le r\}.
$$
By the assumption, $f^2(\Delta_r(X))\subset \Delta_r(Y)$.
If $f$ is not an isometry, there exists $r>0$ such that $f^2(\Delta_r)$ is a proper subset of $\Delta_r$, hence, by compactness, the interior of the complement
$$
\Delta_r(Y) \setminus f^2(\Delta_r(X))
$$
is nonempty, hence, has positive measure. (This is the only place where I am using the manifold assumption.) A contradiction.
The right degree of generality for this proof is that we have two compact path-metric spaces $X, Y$ equipped with Borel measures, each satisfying the property that the measure of each open nonempty subset is positive. The example I gave in the comment shows that this is the right setting.
Edit. This argument works in the case of self-maps, $(X,d_X, \mu_X)=(Y,d_Y,\mu_Y)$. However, in general, it needs more work, as it is unclear why $\Delta_r(X)$ has the same mass as $\Delta_r(Y)$.
A: First, let's prove that $f$ is locally isometric, so that we can focus on small sets, say a closed ball of radius $r\ll 1$ around an arbitrary point. Because $f$ is 1-Lipschitz, the image is also contained in a ball of the same radius so that we have the situation where $f$ is a 1-Lipschitz and volume preserving map $B_r \to B_r$. Since $f(B_r)$ is compact, the complement of the image is open and because of $H_Y^n(B_r \setminus f(B_r)) = (f_\ast H_X^n)(B_r\setminus f_r(B_r) ) = H_X^n(\emptyset)$, it must be empty. Therefore $f$ is surjective.
Second, each boundary point of $B_r$ has exactly one point with distance $2r$, namely the boundary point exactly opposite to it, and all point pairs with distance $2r$ are opposite boundary points. If $y_1=f(x_1), y_2=f(x_2)$ is a pair of opposite boundary points, then $x_1$ and $x_2$ have distance at least $2r$ because $f$ is 1-Lipschitz and are therefore also boundary points. Hence $f$ maps $\partial B_r$ (and only $\partial B_r$) to $\partial B_r$, even respecting opposites, and the interior to the interior. In particular $f$ is open and isometric for points of distance $r$ from the center of the ball. Since $r$ was arbitrary, we conclude that $f$ is a local isometry.
Now let's look at the global case. We already know that $f: X\to Y$ is an open, locally isometric map. In particular it is a local homeomorphism, i.e. a covering map onto its image. By compactness, it must be a covering of finite degree $d$. But again choosing a point and a small enough ball around it, we find $d vol(B) = vol(f^{-1}(B)) = (f_\ast H_X^n)(B) = H_Y^n(B) = vol(B)$ so that $d=1$, i.e. $f$ is injective.
An injective, locally isometric map is a global isometry. QED
A: 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.
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.
