Let
$$
C=\pmatrix{I&0\\0&D},
$$
where the identity $I$ and an arbitrary symmetric $D$ have the same dimension greater than one.
For $v_A=[1,0]^T$ and any nonzero $v_B$ (of the same dimension as $I$ and $D$), $v=v_A\otimes v_B$ is an eigenvector of $C$. The matrix $C$ does not need to have a Kronecker product form since $D$ is arbitrary symmetric.
Short answer: the derivative of $Q\otimes Q\otimes Q\otimes Q$ with respect to $Q$ is a mess, at first sight...
Let's start simple. Let $Q$ be a $K\times K$ matrix with entries $Q_{ij}$ and let $E^{ab}$ be the $K\times K$ matrix with all $0$ entries, except the entry $(a,b)$ which is $1$; in other words, $(E^{ab})_{ij} = \delta_a^i\delta_b^j$.
Then I claim that
$$
\frac{\partial(Q\otimes Q)}{\partial Q_{ij}} = E^{ij}\otimes Q+Q\otimes E^{ij} .
$$
I leave it to you to see why, because trying to write out the involved matrices will probably crash the entire Stack Exchange network...
Jokes aside, this is really immediate to see: just write $Q\times Q$ as in the first formula of the definition and think which elements are affected by $Q_{ij}$. There is the entire $(i,j)$th block, so you get $E^{ij}\otimes Q$, but there is also the $(i,j)$th entry in each block, which gives you $Q\otimes E^{ij}$.
Now, if $A$ and $B$ are matrices which are functions of $Q$, by the same reasoning you get
$$
\frac{\partial(A\otimes B)}{\partial Q_{ij}} = \frac{\partial A}{\partial Q_{ij}}\otimes B + A\otimes \frac{\partial B}{\partial Q_{ij}} .
$$
So you can iterate for instance
$$
\begin{split}
\frac{\partial (Q\otimes Q\otimes Q)}{\partial Q_{ij}}
&= \frac{\partial(Q\otimes Q)}{\partial Q_{ij}}\otimes Q
+ (Q\otimes Q)\otimes \frac{\partial Q}{\partial Q_{ij}} \\
&= (E^{ij}\otimes Q+Q\otimes E^{ij})\otimes Q + (Q\otimes Q)\otimes E^{ij} \\
&= E^{ij}\otimes Q\otimes Q + Q\otimes E^{ij}\otimes Q + Q\otimes Q\otimes E^{ij}.
\end{split}
$$
Now you can prove by induction that
$$
\frac{\partial \bigl(\bigotimes_{n=1}^N Q\bigr)}{\partial Q_{ij}}
= \sum_{n=1}^N \left(\bigotimes_{h=1}^{n-1} Q\right) \otimes E^{ij} \otimes \left(\bigotimes_{h=n+1}^{N} Q\right).
$$
Written more concisely,
$$
\frac{\partial Q^{\otimes N}}{\partial Q_{ij}}
= \sum_{n=1}^N Q^{\otimes (n-1)}\otimes E^{ij} \otimes Q^{\otimes (N-n)} .
$$
Best Answer
Using the well-known Kronecker-vec formula $${\rm vec}(ABC) = (C^T\otimes A)\,{\rm vec}(B)$$ the outer product of two vectors can be expanded in several equivalent ways $$\eqalign{ {\rm vec}(ab^T) &= {\rm vec}(ab^TI_b) = {\rm vec}(I_aab^T) \\ b\otimes a &= (I_b\otimes a)\,b = (b\otimes I_a)\,a \\ }$$ Apply this result to the current the function and calculate the gradient. $$\eqalign{ y &= Mz\otimes{\tt1}_r - {\tt1}_s\otimes p \\ &= (I_m\otimes{\tt1}_r)(Mz) - {\tt1}_s\otimes p \\ dy &= (I_m\otimes{\tt1}_r)M\,dz \\ \frac{\partial y}{\partial z} &= (I_m\otimes{\tt1}_r)M \\ }$$