We have $BGx$ where $B$ is a fixed m×N matrix, and $G$ is an N×n (random) matrix, $x$ is a fixed vector in $R^n$.
By concatenating the rows of $G$ , we can view $G$ as a long vector in $R^{Nn}$.
Consider the linear operator $T$ defined as $T(G) =BGx$ .
Then, Hilbert-Schmidt norm of T equals $||B||_{HS}$ with $||B||^{2}_{HS}$ := $\sum_{i,j} b_{ij}^{2}$
I have tried to write out the identity given by the operator $T$ but I can't quite compare the non $g_{ij}$ term in the resulting vector, i.e. the i_th component of the vectors are as followed
$\sum_{p=1}^{Nn}t_{ip}g'_{p}=\sum_{j=1}^{N}b_{ij}\sum_{k=1}^{n}g_{jk}x_{k}$
with $g'_{p}$ correspond to the j-th component of $G$'s long vector form.
How can I show their HS norm is the same?
*In case my presentation is not clear above, see below for the original content. 
If you see $G$ as a long vector in $R^{Nn}$.
You can use Kronecker product to see the linear operator $T$, defined as $T(G) =BGx$, as $$(x^T\otimes B)vec(G)=[x_1B\,\,x_2B\,\cdots\, x_nB]vec(G).$$ Then, Hilbert-Schmidt norm of T equals $\|x^T\otimes B\|_{HS}$.
You can find related results searching for "(x^T\otimes B) " on SearchOnMath.