Can we arrange $\{1,...,16\}$ in $4\times 4$-grid so {products of rows} = {products of columns}? My motivation is the following question asked by Jacob Steiner, which he then deleted.  For which $n\in\Bbb N$ is it possible to arrange $\{1,\ldots,n^2\}$ in an $n\times n$-grid so that the set of products of columns equals the set of products of rows?
The answer for $n=2$ is clearly No, since the only possibility, up to a transposition and a permutation of rows and columns, is
1 2 
3 4 

Since the set of products of rows are $\{2,12\}$ and the set of products of columns are $\{3,8\}$, this arrangement does not work.  So for $n=2$, it is not possible to make such an arrangement.
 A: RobPratt has answered the question for all $n\le 13$. I shall prove in this answer that when $n\ge 11$, there is no solution.
Let $\pi$ be the prime counting function. I claim is impossible to form a grid whenever
$$
\pi(n^2)-\pi(\lfloor n^2/2\rfloor )>n\tag1
$$
For any $n$ such that the above is true, there are at least $n+1$ primes in the range $\{1,\dots,n^2\}$ which are more than half of $n^2$. This means for any placement of $\{1,\dots,n^2\}$ in an $n\times n$ grid, there will exist two such "big" primes, $p$ and $q$, which are in the same row. The product of that row is a multiple of $pq$. But then none of the column products can be a multiple of $pq$, so no solution exists.
Since $\pi(n)\sim n/\log n$ as $n\to\infty$, by the prime number theorem, the LHS of $(1)$ grows faster than the RHS, so $(1)$ holds when $n$ is sufficiently large. Specifically, I can prove:
Claim: $(1)$ holds whenever $n\ge e^5\approx 148.4$.
My proof uses two numerical inequalities.

*

*For all $n\ge 17$, we have from the  Wikipedia article on $\pi(n)$ that $$\frac{n}{\log n}\le  \pi(n)\le \frac43\cdot  \frac{n}{\log n}$$


*For all $n\ge 10$, we have $$\lfloor n^2/2\rfloor \ge n^2/2-1\ge n^{3/2}.$$
The first inequality is obvious; you can check that the second inequality is true when $n=10$, and by taking the derivative with respect to $n$, you can see that the function $n^2/2-1-n^{3/2}$ is increasing for $n\in [10,\infty)$.
Proof of Claim:
$$
\begin{align}
\pi(n^2)-\pi(\lfloor n^2/2\rfloor )
&\stackrel{1.}\ge \frac{n^2}{\log n^2}-\frac43\cdot \frac{\lfloor n^2/2\rfloor }{\log\lfloor n^2/2\rfloor}
\\&\stackrel{2.}\ge \frac{n^2}{\log n^2}- \frac43\cdot \frac{ n^2/2}{\log n^{3/2}}
=\frac{n^2}{18\log n}
\end{align}
$$
In order to have $\frac{n^2}{18\log n}>n$, it suffices to have $\frac{n}{\log n}>18$. It is easy to check that $\frac{n}{\log n}>18$ holds when $n= e^5$. Since $\frac{n}{\log n}$ is increasing for $n>e$, it holds whenever $n\ge e^5$ as well.
$\square$
Finally, you can check with the help of a computer that $(1)$ holds for all $n$ in the range $[11,148]$. Combined with my claim, this completes the proof that no solution exists when $n\ge 11$. For example, the following Mathematica code returns True:
NoCounterExamples = True;

For[n = 11, n <= 150, n++,
    If[PrimePi[n^2] - PrimePi[Floor[n^2/2]] <= n, 
        NoCounterExamples = False]
];

Print[NoCounterExamples]

A: Not an answer, but pointing out a flaw in the OP logic.
If a prime factor appears singly (i.e. not squared) an odd number of times (like $5$ appearing as $\{5, 10, 15\}$ in the $4\times 4$ grid), that does not imply one of the appearances must be along the diagonal.  E.g.
* 10  *  *
*  * 15  *
5  *  *  *
*  *  *  *

would meet the requirement as far as factors of $5$ are concerned: the first $3$ rows and the first $3$ columns each has exactly one factor of $5$ (and the last row and last column has no factor of $5$).
(These positions represent a (fixed-point-free) permutation in the $3 \times 3$ submatrix.)
UPDATE: Indeed Rob Pratt found such a matrix:
$$
\begin{matrix}   9 &15 &12  &1\\   3 &11 &14  &5\\  6  &7 &13 &16\\  10 &2  &4  &8\end{matrix}
$$
where the positions of the multiples of $5$ represent a fixed-point-free permutation of the $3\times 3$ submatrix after deleting the $3$rd row & column.
So the OP claim that the $4\times 4$ grid must have a factor of $5$ along the diagonal is wrong, and similarly, it is still unproven that $8 \times 8$ cannot be filled because that last $X$ does not need to be any multiple of $11, 17, 19$.  
