Let $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be continuous and monotone, i.e., $$ \left( f(x) - f(y) \right)^\top \left( x-y\right) \geq 0$$ for all $x,y \in \mathbb{R}^n$.
Say for which matrices $A \in \mathbb{R}^{n \times n}$, the function $x \mapsto A f(x)$ is monotone as well.
The following are equivalent:
That $1\implies 2$ is clear. Suppose $1$ is false. Then there is $v\in\mathbb R^n$ such that $Av$ is not a nonnegative multiple of $v$. Define $$f(x) = (v^Tx)v$$ Then $f$ is a monotone map, but $Af$ is not monotone. Indeed, $$(Af(x))^T x =(v^Tx) \, (Av)^Tx \tag{1}$$ Since $Av$ is not a nonnegative multiple of $v$, there is $x$ such that $v^Tx>0$ but $(Av)^Tx<0$. This makes (1) negative.
Of course, for some $f$ there are some non-scalar matrices such that $Af$ is monotone; but what they are depends on what $f$ is. One has to inspect all pairs $(f(x) - f(y), x-y) \in \mathbb R^{2n}$ to find how much perturbation of $f(x) - f(y)$ will preserve monotonicity.