This is idle noodling, and I'm prepared to learn that it's foolish as well as idle. But....

Let $M$ be an $n\times n$ matrix over, oh, let's say an algebraically closed field for now. There have been a few occasions, decades apart, on which I'd have found it very convenient to write the characteristic polynomial of $M$ as $$(X-M)(X-\sigma(M))(X-\sigma^2(M))\ldots(X-\sigma^{n-1}(M))$$ where $\sigma$ is some map of order $n$ from the set of $n$-by-$n$ matrices to itself.

Note that if $n=2$, we can take $\sigma$ to be the classical adjoint of $M$. My main question is:

**Question 1:** For larger $n$, what is the largest set of $n$-by-$n$ matrices on which a reasonably well-behaved $\sigma$ can be defined?

For example, let $B$ be the set of all $n$-by-$n$ matrices with $n$ distinct eigenvalues. Here's an attempt to define $\sigma:B\rightarrow B$.

Let $E$ be the set of all $(M,V_1,V_2,\ldots,V_n)$ such that $M\in B$ and the $V_i$ are the distinct eigenspaces of $M$. Then we can define a map $\tau:E\rightarrow E$ as follows:

Given $(M,V_1,\ldots,V_n)\in B$, let $e_i$ be the eigenvalue $M$ associates to $V_i$. Then set $\tau(M,V_1,\ldots,V_n)=(M',V_1,\ldots,V_n)$ where $M'$ is the matrix having eigenspaces $V_1,V_2,\ldots,V_n$ with associated eigenvalues $e_2,e_3,\ldots,e_n,e_1$.

Then (with $\pi$ the obvious projection) it suffices to find a map $\sigma$ that completes the commutative diagram:

For this, it would suffice to find a section for the map $\pi$, but that looks hopeless even in the case $n=2$, where a map $\sigma$ nevertheless exists. Might $\sigma$ exist for, say, $n=3$? (Note that for $n$ larger than 2, symmetry conditions dictate that $\sigma$ can't be unique.) Or is there perhaps some topological obstruction to this? This leads me to the subsidiary question:

**Question 2:** What is the cohomology of the space $E$?

In case $\sigma$ can be constructed, there are some fairly obvious things to say about extending it to subsets of the $n$-by-$n$ matrices that are strictly larger than $B$, but I'll leave the question here for now.