Of course no. Remember that the operator norm of $A$ wrto the Eucliedan norms is the attained at an eigenvector of $S:=A^TA.$ Try a suitable simple binary $2\times 2$ matrix and compare the values of $\|Ax\|$ on the eigenvectors of $S$ and in the three nonzero binary vectors $(01), (10), (11).$
However, if instead you take in the domain $\mathbb{R}^n$ either the $l^1$ norm $\| \cdot \|\cdot \|1$ |_1$ or the$l^\infty$norm $\| \cdot\|\infty $\|\cdot\|_\infty$ then, whatever norm you have in the target space $\mathbb{R}^m$, the operator norm of $A$ is attained in an extremal point of the unit ball of the domain, which is in both cases a binary vector.
Of course no. Remember that the operator norm of $A$ wrto the Eucliedan norms is the attained at an eigenvector of $S:=A^TA.$ Try a suitable simple binary $2\times 2$ matrix and compare the values of $\|Ax\|$ on the eigenvectors of $S$ and in the three nonzero binary vectors $(01), (10), (11).$
However, if instead you take in the domain $\mathbb{R}^n$ either the $l^1$ norm $\| \cdot \|1$ or the $l^\infty$ norm $\| \cdot\|\infty$ then, whatever norm you have in the target space $\mathbb{R}^m$, the operator norm of $A$ is attained in an extremal point of the unit ball of the domain, which is in both cases a binary vector.