Change of variable within an integral of the Hausdorff measure

In general this formula doesn't hold.

Consider case $n=3$, and $M=\{(x,y,z):0\leq x\leq 1,\quad 0\leq y\leq 1\quad z=0\}$ is a square on $xy$-plane. Define linear transformation $T$ by matrix $$ T_k=\begin{vmatrix}1 && 0 && 0\\0 && 1 && 0\\0 && 0 && k \end{vmatrix} $$ Obviously $\operatorname{det}(T_k)=k$ and $T_k(M)=M$. Take $f:\mathbb{R}^3\to\mathbb{R}$ to be constant function, i.e. $f(x)=1$. Thus we see that $$ \int\limits_{T_k(M)}f(x)H^m(dx)=\iint\limits_{M}1dxdy=1 $$ $$ \operatorname{det}(T_k)\int\limits_{M}f(T_k(x))H^m(T_k(dx))=k\iint\limits_{M}1dxdy=k $$ Hence for $k\neq 1$ $$ \int\limits_{T_k(M)}f(x)H^m(dx)\neq\operatorname{det}(T_k)\int\limits_{M}f(T_k(x))H^m(T_k(dx)) $$ If we take $k=0$ we have counterexample for sigular matrices. If we take $k=2$ we have counterexample for non-singular matrices.