Change of Variables for Hausdorff Measure (Read bounty text for answering question)
Let $H^{m}$ be the $m$-dimensional Hausdorff measure. Let $D$ be a linear transformation matrix. Consider the change of measure formula:
$$
  \int\limits_{A} f(Dx) \; dH^{m}(x) = \int\limits_{ D A} f(y) \; dD_{*}H^{m}(y)
$$
where $D_{*}H^{m}(M) = H^{m}(D^{-1}M)$ is the pushforward of the Hausdorff measure. Is it possible to find such a function $a(x)$ that
$$
   \int\limits_{ D A} f(y) \; dD_{*}H^{m}(y) = \int\limits_{ D A} f(y) a(y) \; dH^{m}(y)
$$
What if we have a self similar object?
$$A= D_1(A) \cup D_2(A)$$
And transform $D_1(A) \rightarrow A$?
 A: Some remarks.  The original question:  

Does a function $a(y)$ exist?  

is (by the Radon-Nikodym theorem) equivalent to the question  

Is the image measure $D_*H^m$ absolutely continuous with respect to $H^m$?  

(Let's suppose $A$ has sigma-finite measure.)  And the answer is yes if $D$ is a non-singular matrix; the answer is sometimes no if $D$ is a singular matrix.  But even knowing function $a(y)$ exists doesn't satisfy Zach, who wants an explicit formula.
Example (for the singular case)  
Consider the Cantor singular function

Our set $A$ is the graph of the function.  It is a set with arc length exactly $2$ (that is, $1$-dimensional Hausdorff measure exactly $2$.)  
Take $D$ to be the projection onto the $x$-axis.   What is the image measure $D_*H^1$?  Images of all the horizontal line-segments give us exactly Lebesgue measure in the real line.  But images of the rest (a set of $H^1$-measure $1$) give us a singular measure concentrated on the Cantor set.  The image measure is not absolutely continuous.  
Or, alternatively, take $D$ to be the projection onto the $y$-axis.  Now the line segments map to countably many point-masses (total mass $1$), while the rest of the graph maps exactly onto Lebesgue measure on the $y$-axis.  But again, the image measure is not absolutely continuous.
There is a general theorem (due to Besicovitch) that generalizes this example.  A set $E$ in Euclidean space with $0 < H^1(E) < \infty$ may be written as a disjoint union of two parts: one part is (up to measure zero) a countable union of rectifiable curves, the other part is "curve-free" and "dust-like".  (See K. J. Falconer, Fractal Geometry, 3rd edition, section 5.2: "Structure of $1$-sets")
Remark (non-singular case:   $D_*H^m \ll H^m$)  
Suppose the matrix is non-singular.  Then we have inequalities of the form
$$
\alpha \; \text{diam}\; A \le
\text{diam}\; D(A) \le
\beta \; \text{diam}\; A
$$
for all sets $A$.  Here $\alpha, \beta$ are the smallest and largest singular values of $D$, and they are positive because $D$ is nonsingular.
Using that inequality we get
$$
\alpha^m H^m(A) \le H^m(D(A)) \le \beta^m H^m(A),
\\
\alpha^m H^m(D^{-1}(A)) \le H^m(A) \le \beta^m H^m(D^{-1}(A)),
$$
so $D_*H^m \ll H^m$ as claimed.
