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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 $J(Af) = AJ(f)$?

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    yes. If $A$ is constant and we identify $A$ with the linear transformation it induces. Otherwise, $Af$ is not a "composite" function, it's the function defined by matrix multiplication of $f$ on $A$. $A$ being constant, it pulls out. – 2017-02-18
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    @JamesS.Cook Thanks for your work in complexity theory – 2017-02-18
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    Which work is that? (maybe I'm not the same James Cook...) – 2017-02-18

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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.