I have to proove that $(K_{p,p}⊗ I_p)(\vec{a}⊗\vec{a}⊗\vec{a})$ equals $(\vec{a}⊗\vec{a}⊗\vec{a})$, where $\vec{a}$ is px1 vector, $I_p$ is unit matrix pxp and $K_{p,p}$ is commutation matrix ppxpp. I have some ideas, but I'm not sure, do I think right. I tried to proove it from right to left. Starting from $(\vec{a}⊗\vec{a}⊗\vec{a})$ = $((\vec{a}⊗\vec{a})⊗\vec{a})$
I used to fact, that $(\vec{a}⊗\vec{a})= K_{p,p} \cdot(\vec{a}⊗\vec{a}) \cdot K_{1,1}$. And I know, that commutation matrix is orthogonal, so $K_{p,p}'\cdot K_{p,p}=I_{p,p}$. So I have $K_{p,p}\cdot I_{p,p}\cdot (\vec{a}⊗\vec{a})$. But are $K_{p,p}\cdot I_{p,p}$ and $K_{p,p}⊗I_{p}$ equal?
I would like to give a direct proof for the particular case $p=2$ involving matrices with dimension $p^3=8$.
Let $\mathbf{a}=\pmatrix{p\\q}$.
We have
$$b:=\mathbf{a}\otimes \mathbf{a}=\pmatrix{p^2\\pq\\pq\\q^2} \ \text{and} \ c:=(\mathbf{a}\otimes \mathbf{a})\otimes \mathbf{a}= \pmatrix{p^3\\p^2q\\p^2q\\pq^2\\p^2q\\pq^2\\pq^2\\q^3}$$
Let use apply property $$ (\mathbf{A} \otimes \mathbf{B})(\mathbf{C} \otimes \mathbf{D}) = (\mathbf{AC}) \otimes (\mathbf{BD})$$
(That you can find for example in the Wikipedia article about Kronecker product) to :
$$ (\mathbf{K_{2,2}} \otimes \mathbf{I_2})((\mathbf{a}\otimes \mathbf{a}) \otimes \mathbf{a}) = (\mathbf{\mathbf{K_{2,2}} b}) \otimes (\mathbf{I_2a})$$
which can be transformed into :
$$\pmatrix{1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1}\pmatrix{p^2\\pq\\pq\\q^2}\otimes \mathbf{a}$$
which indeed is equal to :
$$\pmatrix{p^2\\pq\\pq\\q^2}\otimes \mathbf{a}= (\mathbf{a}\otimes \mathbf{a}) \otimes \mathbf{a}$$
as desired.
Edit :
What have we learned from this particular example ? Look at the last relationship : the key fact (independent of dimension $p$) is that :
$$K_{p,p} (a \otimes a)= a \otimes a $$
which remains to be proven in the general case, but which is a lot simpler than the initial question.