Let $\mathbf{X}\in \mathbb{R}^{n\times d}$ be a matrix with orthonormal columns ($d\le n$). By sub-multiplicativity, we have that $\|\mathbf{A} \mathbf{X}\|_2 \le \|\mathbf{A}\|_2$ and $\|\mathbf{X} \mathbf{B}\|_2 \le \|\mathbf{B}\|_2$ for arbitrary matrices $\mathbf{A}\in \mathbb{R}^{m\times n}$ and $\mathbf{B}\in \mathbb{R}^{d\times p}$. In either case, does equality hold true in general?
I think in general $\|\mathbf{X} \mathbf{B}\|_2 = \|\mathbf{B}\|_2$ holds because $\|\mathbf{X} \mathbf{B}\|_2^2 = \|\mathbf{B}^\top \mathbf{X}^\top \mathbf{X} \mathbf{B}\|_2 = \|\mathbf{B}^\top \mathbf{B}\| = \|\mathbf{B}\|_2^2$ but not the other case.
No it is not true in general. For example, let $d=1,m=1$, $n=2$ and $A = [0, 1]$, $X = [1,0]^T$. Then we have $$ \|AX\|_2 = 0 < \|A\|_2 = 1.$$