For some field $F$, given a set of $n$-dimensional input vectors $V_{in}$ and a set of $n$-dimensional output vectors $V_{out}$, where each of the set does not contain any duplicated elements.
Is it possible to create a corresponding linear transformation matrix $M$ such that $\vec{v}_{out} = M \vec{v}_{in} $ where $\vec{v}_{in} \in V_{in}, \vec{v}_{out} \in V_{out}$?
$M:F^n \rightarrow F^n$
On
Let $V_{\text{in}}$ be an ordered set of $n$ vectors of $F^n$. Then the assertion “for every ordered set $V_{\text{out}}$ of $n$ vectors of $F^n$, there is a linear transformation $T\colon F^n\longrightarrow F^n$ such that $T$ maps the $k$th element of $V_{\text{in}}$ into the $k$th element of $V_{\text{out}}$” is equivalent to the assertion that $V_{\text{in}}$ is a basis of $F^n$.
This is not true in general. Suppose we are in $ \mathbb{R}^2 $. Let $ V_{\text{in}} = \{ (1, 0), (2, 0) \} $, $ V_{\text{out}} = \{ (1, 0) \} $. Then if a linear transformation $ T $ satisfied this property, we would have $ T((1, 0)) = T((2, 0)) = (1, 0) $. Then by linearity, we would also have $ T((1, 0)) = T((2, 0)) - T((1, 0)) = (0, 0) $, which is a contradiction.