This might sound quite trivial. Let $\mathbf{x} = \begin{pmatrix}x_1,\ldots,x_N \end{pmatrix}^\top$ which takes on values on the $N$-dimensional orthant $\mathcal{O}_N = [0,\infty)^N$. If $\mathbf{y}$ is linear transformation of $\mathbf{x}$ defined as $\mathbf{y} = \mathbf{A}\mathbf{x}$ where $\mathbf{A}$ is an $N\times N$ transformation matrix (i.e., invertible),
$\mathbf{1.}$ is the image still $\mathcal{O}_N$?
$\mathbf{2.}$ If no, is there a necessary and sufficient condition in $\mathbf{A}$ for the the image to be preserved?
Context: Currently working on multivariate integration involving change-of-variables resulting to change in domain of integration.
No, in general the image is not $\mathcal{O}_N$. This is easy to see, for example take $N=2$ and $A$ to be the standard rotation matrix: $$\begin{bmatrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \\ \end{bmatrix} $$
Here, if you put $\theta = 90^{\circ}$, then the image is completely disjoint from $\mathcal{O}_N$. Or you can use the shear matrix to obtain situations where the image is strictly contained in $\mathcal{O}_N$ or will strictly contain $\mathcal{O}_N$.
For your second question, it is easy to detect if the image is contained in $\mathcal{O}_N$. To check that, we just need to check if all the basis vectors of $\mathcal{O}_N$ land in $\mathcal{O}_N$.
To check if the image is exactly $\mathcal{O}_N$, we'll have to check whether every element of $\mathcal{O}_N$ (or at least the basis) has a preimage which will be no different than looking at the image of $\mathcal{O}_N$ and comparing.