What is the Jacobian of $A*f$? Let $f: \mathbb{R}^n \to \mathbb{R}^n$
Let $A$ be a real matrix of dimension $n \times n$
Consider the composite function $$g = Af$$
Question: Let $J(f)$ be the Jacobian of $f$, then does $J(Af) = AJ(f)$?
 A: Suppose $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is differentiable at $a \in \mathbb{R}^n$ then there exists $d_af: \mathbb{R}^n \rightarrow \mathbb{R}^n$ which approximates the change in $f$ at $a$ in the sense that:
$$ \frac{f(a+h)-f(a)-d_af(h)}{\| h \|} =\eta_f(a,h) \rightarrow 0$$
as $h \rightarrow 0$. Furthermore, $d_af$ is a linear transformation hence its action is captured via matrix multiplication; $d_af(h) = J_f(a)h$. Explicitly: $J_f$ is the $n \times n$ matrix: 
$$ J_f = \left[ \frac{\partial f}{\partial x_1} \bigg{|}\frac{\partial f}{\partial x_2}\bigg{|} \cdots \bigg{|} \frac{\partial f}{\partial x_n} \right] $$
Consider the function $g(x) = Af(x)$ where this formula is matrix column multiplication by the constant square matrix $A$. Observe,
$$ g(a+h)-g(a) = Af(a+h) - Af(a) = A[f(a+h)-f(a)] $$
since $f(a+h)-f(a) = d_af(h)+\| h \| \eta_f(a,h)$ we have
$$ g(a+h)-g(a) = A[d_af(h)+\| h \| \eta_f(a,h)]$$
thus
$$ \frac{g(a+h)-g(a) -Ad_af(h)}{\| h \|} =  A\eta_f(a,h) $$
then by continuity of matrix multiplication (you can work out the epsilonics by using the maximum entry in $A$) we find $A\eta_f(a,h) \rightarrow 0$ as $h \rightarrow 0$ and we conclude the linear transformation $Ad_af$ is the differential of $g$ at $a$. In particular,
$$ Ad_af(h) = AJ_f(a)h \ \ \Rightarrow \ \ J_g(a) =AJ_f(a) $$
but, $a$ is arbitrary, so $J_g = AJ_f$ or $J(Af) = AJ(f)$ as you wished. 
Alternatively, define $T(x)=Ax$ and observe $(T \circ f)(x) = Af(x)$. Moreover, $d_a(T \circ f) = d_{f(a)}T \circ d_af$ hence $J_{T\circ f}(a) = J_T(f(a))J_f(a)$ but $J_T=A$ since $T$ is linear hence $J_{T\circ f}(a) = AJ_f(a)$ as you wished.
So, either by direct computation or via the chain rule of advanced calculus we can see the result you wish.
