Implications of $\|Ax\|_1\leq \|Bx\|_1$

Implications of $\|Ax\|_1\leq \|Bx\|_1$

Let us assume that $A,B\in\mathbb{R}^{m\times n}$ and $\|\cdot\|_1$ is the
(vector) norm-1 defined as $\|x\|_1=\sum_{i=1}^n|x(i)|$, and the induced
matrix norm is $\|A\|_1=\sup_{\|x\|_1=1}\|Ax\|_1$. Let us assume that:
$$ \|Ax\|_1 \leq \|Bx\|_1, \forall x\in\mathbb{R}^n. $$
What does this imply about $A$ and $B$? I can only tell that if $n=m$ and
$B$ is invertible then we may set $y=Bx$ so $x=B^{-1}y$, hence the above
inequality becomes:
$$ \|AB^{-1}y\|_1 \leq \|y\|_1, \forall x\in\mathbb{R}^n, $$
Therefore,
$$ \|AB^{-1}\|_{1}\leq 1 $$
But, I would be interested in the case where $B$ is not assumed to be
invertible and may also not be a square matrix.