Let $\lVert \cdot \rVert_a$ and $\lVert \cdot \rVert_b$ denote the $\ell_a$ and $\ell_b$ norms on $\mathbb{R}^n$ ($a,b$ are positive integers or $\infty$, $a \neq b$). Let $M_n(\mathbb{R})$ be the space of $n \times n$ real matrices.
Now suppose $A \in M_n(\mathbb{R})$ is a contraction on $\mathbb{R}^n$ equipped with the metric induced by $\lVert \cdot \rVert_a$, i.e. $$ \exists c \in (0,1) \qquad \text{ s.t. } \qquad \lVert Ax - Ay \rVert_a \leq c \lVert x-y \rVert_a \qquad \forall x,y \in \mathbb{R}^n. $$
Is it true that $A$ is also a contraction under the metric induced by $\lVert \cdot \rVert_b$? If not, it would be great if a counterexample is given.
I would also like to know if there are any 'specific' choices of $a$ and $b$ for which the answer to the above question is yes.
Take $a=2, b=\infty$ and $A=\pmatrix{1/2&1/2\\0&0}$. Then $\lVert Ax\rVert_a\leq(1/\sqrt2 )\lVert x\rVert_a$ for all $x$, but $\lVert A \pmatrix{1\\1}\rVert_b=1 \lVert\pmatrix{1\\1}\rVert_b$.
Remark: The first relation holds true, since with $x=\pmatrix{x_1\\x_2}$ we have $Ax=(1/2)\pmatrix{x_1+x_2\\0}$ and since moreover $(x_1+x_2)^2\leq 2(x_1^2+x_2^2)$.