The metric can not be recovered by its Hausdorff measure in generally. 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 $f_*\mathcal{H}^n_X(X)=\mathcal{H}^n_Y(Y)$.

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 $f_*\mathcal{H}^n_X(X)=\mathcal{H}^n_Y(Y)$.

The metric can not be recovered by its Hausdorff measure in generally. Now, assume that $(X,d_X)$ and $(Y, d_Y)$ are connected compact length spaces and induce $n$-dimensional Hausdorff measure $\mathcal{H}^n_X$ 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?

The metric can not be recovered by its Hausdorff measure in generally. 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 $f_*\mathcal{H}^n_X(X)=\mathcal{H}^n_Y(Y)$.