Let $W_{i,1},W_{i,2},W_{i,3} \in \mathbb{R}^{n \times n}$, $i \in {1,2}$ be such that
$$ \eqalign{ \Big[\matrix{W_{i,1}^T & W_{i,2}^T & W_{i,3}^T}\Big] \left[\matrix{W_{i,1}\\W_{i,2}\\W_{i,3}}\right] = I } $$
where $I$ is the identity matrix. Now, let
$$\eqalign{ W = \left[\matrix{ W_{2,1}W_{1,1} \\ W_{2,2}W_{1,1} \\ W_{2,3}W_{1,1} \\ W_{1,2} \\ W_{1,3}} \right] }$$
Then, $W^{T}W = I$ but I don't see why. How can this idea be expanded towards the Kronecker products, i.e., $W_{i,j} \otimes W_{i,k}$ for $i \in \{1,2\}$ and $j,k \in \{1,2,3\}$?
Let's start with showing $W^\top W = I$. We can write the assumption about the $W_{i,j}$ as $$ \begin{align} \label{eq:condition}\tag{\(*\)} W_{i,1}^\top W_{i,1} + W_{i,2}^\top W_{i,2} + W_{i,3}^\top W_{i,3} = I, \qquad i = 1,2. \end{align} $$
Now let's expand $W^\top W$ in a similar way.
$$ \begin{align*} W^\top W &= \Big[\begin{array}{ccccc} (W_{2,1}W_{1,1})^\top & (W_{2,2}W_{1,1})^\top & (W_{2,3}W_{1,1})^\top & W_{1,2}^\top & W_{1,3}^\top \end{array}\Big] \left[\begin{array}{c} W_{2,1}W_{1,1} \\ W_{2,2}W_{1,1} \\ W_{2,3}W_{1,1} \\ W_{1,2} \\ W_{1,3} \end{array}\right] \\ &= W_{1,1}^\top W_{2,1}^\top W_{2,1} W_{1,1} + W_{1,1}^\top W_{2,2}^\top W_{2,2} W_{1,1} + W_{1,1}^\top W_{2,3}^\top W_{2,3} W_{1,1} + W_{1,2}^\top W_{1,2} + W_{1,3}^\top W_{1,3} \\ &= W_{1,1}^\top \left(W_{2,1}^\top W_{2,1} + W_{2,2}^\top W_{2,2} + W_{2,3}^\top W_{2,3} \right)W_{1,1} + W_{1,2}^\top W_{1,2} + W_{1,3}^\top W_{1,3} \end{align*} $$ Now recognize that the sum in the parentheses is \eqref{eq:condition} with $i = 2$. Then we have a simplification: $$ \begin{align*} W^\top W &= W_{1,1}^\top (I) W_{1,1} + W_{1,2}^\top W_{1,2} + W_{1,3}^\top W_{1,3} \\ &= W_{1,1}^\top W_{1,1} + W_{1,2}^\top W_{1,2} + W_{1,3}^\top W_{1,3} \\ &= I, \end{align*} $$ using \eqref{eq:condition} again but with $i=1$.
Not sure how this relates to the Kronecker product though, especially since $W$ has matrix-multiplied blocks instead of element-wise multiplications. Note though that $W$ has orthonormal columns, since $W^\top W = I$, but the $W_{i,j}$ don't necessarily, because $W_{i,j}^\top W_{i,j} = I$ for all $i,j$ would contradict \eqref{eq:condition}.