If $q\neq 0$ and $e$ are column vectors, when do we have $A$ such that $Aq=e$? Question: if $q\neq 0$ and $e$ are $n\times 1$ column vectors, when do we have $A$ ($n\times n$) such that $Aq=e$?
I've got a feeling that such $A$ always exists. Let $q_j$ be a nonzero element of $q$ and construct the $i$-th row of $A$ as ($i=1,\ldots,n$):
$$
A_i'=\begin{pmatrix}0&\ldots&\underbrace{\frac{c_i}{q_j}}_{j\text{-th position}}&\ldots&0\end{pmatrix}
$$
and let
$$
A=\begin{pmatrix}A_1'\\ \vdots \\ A_n'\end{pmatrix}\cdot
$$
Did I miss something? I've been making a lot of mistakes all morning so my confidence is shaky.

The matrix you constructed does precisely that, but you don't only want $q\ne 0$ (as in $q$ is not the $0$-vector) but you want $q_i \ne 0\ \forall i$ (as in $q$ has no zero entries).
A more compact way to write this would be
$$A = \mathrm{diag}(e \div q)$$
Where $\div$ denotes component-wise division.
Another option is
$$A = \frac1{\|q\|_2^2} eq^T = \frac{eq^T}{q^Tq}$$
