Can you determine any number of magic squares, that when treated as matrices, can be applied mathematical operations to return a new magic-matrix?
You cannot use the same matrix twice!
The answer should be given as a mathematical equation using the matrices you find.
Magic squares follow the general magic square rules, they contain 1, .., $n \times n$ once, and each row, column, and diagonal sum to 1 number.
Allowed Operations: Add, Subtract, Multiply, Exponention, Logarithms
It is possible.
Fix $n$. Let $M$ denote the set of all $n$-by-$n$ magic squares. $M$ is really huge - for $n=6$, we have $|M| \sim 10^{19}$ (see OEIS).
Consider all $2^{|M|}$ sums of subsets of $M$. Each one is an $n$-by-$n$ matrix whose entries are positive integers which are at most $n^2|M|$. There are $(n^2|M|)^{n^2}$ matrices which fit this description.
If $|M|$ is large enough, then $2^{|M|} > (n^2|M|)^{n^2}$. Then by Pigeonhole, two of the sums of subsets are equal. Using this equality, we can express one of the magic squares as a sum/difference of others.
What is "large enough"? Taking logs on the inequality, we find that even $|M|=O(n^3)$ is large enough for sufficiently large $n$. For $n=4,5,6$ this works. I cannot find any references to back this up, but I bet there are enough magic squares that this works for all $n\geq4$.
If we consider two magic squares that differ by a rotation distinct, then here's an example with $3\times 3$ squares. Let $$ M_1=\left[\begin{array}{ccc}4&9&2\\3&5&7\\8&1&6\end{array}\right],\,M_2=\left[\begin{array}{ccc}8&3&4\\1&5&9\\6&7&2\end{array}\right],\,M_3=\left[\begin{array}{ccc}6&1&8\\7&5&3\\2&9&4\end{array}\right],M_4=\left[\begin{array}{ccc}2&7&6\\9&5&1\\4&3&8\end{array}\right]. $$ Then we can compute $$ M_1+M_3=M_2+M_4=\left[\begin{array}{ccc}10&10&10\\10&10&10\\10&10&10\end{array}\right], $$ so that the equation $M_1+M_3-M_2=M_4$ gives a solution to the problem.
