How to transform $z$ into $\hat z$ Assume that we have vector $a = \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix} \in \mathbb R^n$, $b = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix} \in \mathbb R^n$, $c = \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix} \in \mathbb R^n$.
Then, the vector $z = \begin{bmatrix} a \\ b \\ c \end{bmatrix}\in \mathbb R^{3n}$.
How to obtain the vector $\hat z = \begin{bmatrix} a \\ a  \\ b \\ b \\ c \\ c \end{bmatrix}\in \mathbb R^{2\times 3n}$ based on $z$?
The Kronecker product is not useful in this case, because it focuses on elements, instead of subvectors.

I believe the Kronecker product can be pretty much useful. You could take.
Approach 1: Take:
$$z_{1} = \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\0  \\ 0\end{pmatrix} \quad z_{2} = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} \quad z_{3}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\1 \end{pmatrix} $$
so that:
$$\hat{z} = z_{1}\otimes a + z_{2}\otimes b + z_{3}\otimes c$$
Approach 2: Take:
$$z_{1} = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} \quad z_{2} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \end{pmatrix} \quad z_{3} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix}$$
so that:
$$z'=\begin{pmatrix}  a & b & c \\ a & b & c \end{pmatrix} = z_{1}\otimes a + z_{2}\otimes b + z_{3}\otimes c$$
and take $\hat{z}$ to be the vectorization of $z'$.
