About norm equivalence of vectors Let $x$ be vector in $\mathbb R^n$, and let $\| \cdot\|_a$ and $\| \cdot\|_b$ be two norms on $\mathbb R^n$. We know that norms are equivalent on $\mathbb R^n$, so there are $L, M >0$ such that
$$
L \| x\|_a \leq \| x\|_b \leq  M \| x\|_a.\\
$$
Can we say that there exists matrices $A$ and $B$ such that $\| x\|_b = \| Ax\|_a$ and $\| x\|_a = \| Bx\|_b$?
 A: Not always; it can depend on the geometry of the unit balls of each norm. For example, the Euclidean norm is strictly convex, meaning that
$$\left\|\frac{x + y}{2}\right\|_2 < \frac{\|x\|_2 + \|y\|_2}{2},$$
whenever $x \neq y$. Note the strict inequality here!
(This is equivalent to the condition that the interior of the line segment between any pair of distinct points in the ball, lies in the interior of the ball, or equivalently, the sphere contains no line segments.)
This property fails to hold for the $\|\cdot\|_1$ norm; if $e_i, e_j$ are two distinct standard basis vectors of $\Bbb{R}^n$, then
$$\left\|\frac{e_i + e_j}{2}\right\|_1 = 1 = \frac{\|e_i\|_1 + \|e_j\|_1}{2}.$$
(Again, note now that the line segment between $e_i$ and $ej$ lies in the sphere of the ball, not the interior, so we don't have strict convexity.)
Now, if there were a matrix $M$ such that $\|Mx\|_2 = \|x\|_1$, then
\begin{align*}
\left\|\frac{Me_i + Me_j}{2}\right\|_2 &= \left\|M\left(\frac{e_i + e_j}{2}\right)\right\|_2 \\
&= \left\|\frac{e_i + e_j}{2}\right\|_1 \\
&= \frac{\|e_i\|_1 + \|e_j\|_1}{2} \\
&= \frac{\|Me_i\|_2 + \|Me_j\|_2}{2}.
\end{align*}
If $Me_i \neq Me_j$, then this would contradict the strict convexity of $\|\cdot\|_2$, thus we must have $Me_i = Me_j$. That is, $M$ maps all the standard basis vectors to a single vector!
I'm guessing you were implicitly assuming that $M$ was invertible (as I did, before I edited this answer!). We can still continue without this assumption. The above condition makes $M$ rank $1$ (note $M$ cannot be rank $0$, otherwise $\|x\|_1 = 0$ for all $x$, which is absurd). So, there exists some non-zero $v$ such that
$$M(a_1e_1 + \ldots + a_n e_n) = (a_1 + \ldots + a_n)v.$$
So,
\begin{align*}
|a_1| + \ldots + |a_n| &= \|a_1e_1 + \ldots + a_n e_n\|_1 \\
&= \|M(a_1e_1 + \ldots + a_n e_n)\|_2 \\
&= \|(a_1 + \ldots + a_n)v\|_2 \\
&= |a_1 + \ldots + a_n|\|v\|_2.
\end{align*}
If we consider the vector $(1, -1, 0, 0, \ldots, 0) \in \Bbb{R}^n$, then the left hand side is $2$, while the right hand side is $0$, giving us our final contradiction.
