Prove $|\det(J_{\phi}(a))|=k$ where $v(\phi (A))=kv(A)$ and $A$ is a Jordan Measurable set. 
$(i)$ Let $\phi: \mathbb R^n \to \mathbb R^n$ be continuously differentiable homeomorphism such that $J_{\phi}(a)$ is invertible matrix for every $a \in \mathbb R^n$.
Suppose there is a constant $k>0$ such that for every Jordan Measurable set A,
$v(\phi (A))=kv(A)$.
Prove $|\det(J_{\phi}(a))|=k$ for all $a \in \mathbb R^n$

I am completely stuck on this. I considered using the Change of Variable Theorem  or the fact that I know:
If $A$ is a bounded, Jordan measurable subset of $\mathbb R^n$, then given $c ∈ R$, the set
$B := \{(x_1, \dots , x_{n−1}, cx_n) : (x_1, \dots , x_n) \in A\}$
is Jordan measurable and $v(B) = |c|v(A)$
or that:
If $T \ M_n(\mathbb R)$ is a matrix and $ A \subset \mathbb R^n$
is a
bounded, Jordan measurable function, then
$T A := \{T x : x \in A\}$
is Jordan measurable and
$v(T A) = |\det T|v(A)$
but I keep tying myself in knots.
Could anyone provide a simple, clear way to prove these?

Surely you know:
Lemma: If $f,g$ are integrable and $\int_{A}f=\int_{A}g$ for every measurable set $A$, then $f=g$ almost everywere.
Then,
by the change of variables theorem: 
$$ \int_{A}|det(J_{\phi}(x)|dx =\int_{\phi(A)}dx=v(\phi(A))=kv(A)=\int_{A}kdx$$
for every (Jordan)measurable set $A$. Then, $|det(J_{\phi}(x)|=k$ almost everywere. Since $|det(J_{\phi})|$ is continuous, the last equality holds for every $x\in\mathbb{R}^{n}$.
