Minimize matrix distance to tensor product Minimize the following function:
$ f(V) = || V \otimes V - U_1 \otimes U_2 ||$
where $U_1, U_2 \in SU(n)$ are fixed and we minimize over all $V \in SU(n)$. The norm is from the trace inner product.
The minimum value of $f$ and the $V$ which minimizes it are both important.
Cross posted on mse.
 A: I was hoping someone else would do this, because, although it's not difficult, it's complicated. Here is a qualitative version. Unless $U$ is itself relatively close to $I$ (this will be formulated properly), the minimum will be close to $2$. For example, if $U$ has eigenvalues the $n$ $n$th roots of unity, then the minimum is approximately $2- 1/n$. 
Again (cf mathoverflow.net/questions/244405/), write $V \otimes V - I \otimes U = (V \otimes VU^{-1}) (I \otimes U)$. It suffices to minimize the distance from $1$ of all the eigenvalues of $V \otimes VU^{-1}$.
It is convenient to also measure the distance between two points on the unit circle via the length of the shorter arc joining them (given values in radians, with a maximum of $\pi$). 
The eigenvalues of $V \otimes VU^{-1}$ are exactly the products $a\cdot b$ where $a$ is an eigenvalue of $V$ and $b$ is an eigenvalue of $VU^{-1}$. 
Associate to a finite set of elements of the unit circle, the shortest arc on the unit circle that contains all elements of the set (if the set is a singleton, this is just a point; if the set contains two points, there may be two choices if one of them is minus the other, in which case, it really doesn't matter which arc we take; with three or more points, there is no ambiguity). We also consider the mid-point of the arc (meaning the point on the arc equidistant from both endpoints; with this definition of mid-point, we always get a point of modulus one).
Begin with the spectrum of $VU^{-1}$; construct the arc and its mid-point. Say the length of the arc is $t$ radians and its midpoint is $z = e^{ i \theta}$ (we can assume $0 \leq\theta < \pi$ by replacing $V$ by $-V$ if necessary). This yields $\| V U^{-1} - zI\| = 2 \sin t/4$. Similarly, we have the arc of length $u$ with mid-point $y = e^{  i \psi}$ for the spectrum of $V$, and correspondingly, $\| V - yI\| = 2 \sin u/4$.
The eigenvalues of the tensor product are all possible produces of the spectra of each; the four products arising from the endpoints will yield an arc that either covers the entire circle (only possible if $u = t = \pi$), or has length exactly $u + t$ radians, and is centred at $yz$. Then the distance from the identity of $V \otimes VU^{-1}$, that is, the norm of $\| V \otimes VU^{-1}\| $  is at least $ 2 \sin (u+t)/4 $, and then equality holds only if $yz = 1$. If this is small, then even smaller is $2\sin t/4$. But this forces $\| V - zU\| = 2 \sin t/4$; combined with $\| V - yI \| = 2 \sin u/4$, we have $\|zU - yI\| \leq 2 (\sin u/4 + \sin t/4) $, which means the spectrum of $U$ is contained in arc of length roughly (when $u + t$ is small) $u+t$ with midpoint $z^{-1}y$. This is what I meant when I said $U$ has to be close to $1$ (if the minimum norm is small).
In this argument, we've only used that $U$ is unitary, not that it has determinant one; so we also obtain a condition on the $y$ and $z$ in $\|zU - yI\| \leq 2 (\sin u/4 + \sin t/4) $ (unfortunately, not a very exciting one, since we have to take $n$th powers ...). Since $V$ and $VU^{-1}$ are also supposed to be in SU($n$), there are corresponding conditions arising from $\| V - yI\| = 2\sin u/4$, and from $\| V U^{-1} - zI\| = 2 \sin t/4$. For example, $|y^n -1| \leq 2n\sin u/4$.
The remaining case occurs if $u = t = \pi$, and that can be handled directly; this results in the norm equalling $2$, which is the maximum in any case. 
The upshot qualitatively is that the minimum value is only going to be small if $U$ is already close to a scalar matrix. The quantitative version doesn't look difficult, just boring.
