Is a constant Jacobian an equivalent condition to linearity? And furthermore, does a constant, non-zero Jacobian imply that the function is invertible?
 A: I presume you're talking about the Jacobian matrix, rather than the Jacobian determinant (but see the Jacobian conjecture).
If differentiable function 
$F: \mathbb R^m \to \mathbb R^n$ has constant Jacobian matrix $J$, then
$F(b) - F(a) = \int_\Gamma J \; dx = J \; (b - a)$ where $\Gamma$ is the straight line segment from $a$ to $b$.  This says $F$ is affine: $F(x) = F(0) + J \; x$.  However, 
there's no reason for $F(0)$ to be $0$, so it's not necessarily linear.
If in addition $J$ is invertible (not just nonzero), then multiplication by $J$ is
an invertible function.
A: For your first question, a constant Jacobian does not necessarily mean that the function is linear. 
For your second question, you don't need the Jacobian to be constant, you just need it to be non-zero. The inverse function theorem states that for $E,F$ are normed linear spaces, for some open $U\subseteq E$, $x_0\in U$, and $f:U\to F$ is a $C^1$ map, that is it's differentiable and the derivative, as a linear map, is continuous; if $f'(x_0):E\to F$ is invertible then $f$ is locally $C^1$ invertible at $x_0$, and if $\varphi$ is its local inverse, and $y=f(x)$, then $\varphi'(y)=f'(x)^{-1}$. 
For a concrete example, let $f,g\in C^1$, and let $\varphi:\mathbb{R}^2\to\mathbb{R}^2$ be given by $$ \varphi(x,y)=(x+x^2f(x,y),y+y^2g(x,y))$$
After some calculation, you can find that the Jacobian of $\varphi$ at $(0,0)$ is:
 $$ [\varphi'(0,0)] = \begin{pmatrix}
1 & 0 \\
0 & 1\\
\end{pmatrix}$$ 
Telling us that $\varphi$ is locally $C^1$ invertible at $(0,0)$, which should make sense by inspection of $\varphi$, one notices that it is a perturbation of the identity map by those second order terms. In a similar light, consider the vector valued function $h:\mathbb{R}^2\to\mathbb{R}^2$ defined by 
$$ h(x,y)=\begin{pmatrix}
e^x\cos y\\
e^x\sin y\\
\end{pmatrix}$$
The Jacobian matrix is:
$$J_h(x,y) =  \begin{pmatrix}
e^x\cos y & -e^x\sin y \\
e^x\sin y & e^x\cos y\\
\end{pmatrix}$$
with determinant $\det J_h(x,y)=e^{2x}\cos y+e^{2x}\sin y=e^{2x}\neq 0$ everywhere. So by the inverse function theorem, for every point $z\in \mathbb{R}^2$, there exists a neighborhood about $z$ over which $h$ is invertible, but this doesn't mean $h$ is invertible over it's entire domain since it isn't even injective $(h(x,y)=h(x,y+2\pi))$.
