When is the trace multiplicative? Let $A,B\in M_n(\mathbb{R})$ be real (or complex) square matrices. Generally speaking,
$${\rm trace}(AB)\neq {\rm trace}(A){\rm trace}(B)$$
There are a lot of 'easy' examples where this doesn't hold. What is not obvious to me is whether there is any chance of it working for something other than $A=B=0$, the zero matrix. Even if you work out the details for $n=2$ the algebraic condition is not very insightful.
Question: What are sufficient or necessary conditions on $A$ and $B$ under which the trace is multiplicative, i.e., ${\rm trace}(AB)= {\rm trace}(A){\rm trace}(B)$?
Edit: This post from 2014 asked whether the trace is multiplicative or not. I am well aware that it is not, but a distinct question is whether there are conditions on $A$ and $B$ under which the trace becomes multiplicative.
 A: Not really an answer, but too long for a comment.
Consider just diagonal matrices. Let
\begin{align}
A =
\begin{bmatrix}
a & 0\\
0 & a
\end{bmatrix} \ \ \text{ and } \ \ 
B =
\begin{bmatrix}
b & 0\\
0 & b
\end{bmatrix},
\end{align}
then
\begin{align}
\operatorname{tr}(AB) = 2ab = 4ab= \operatorname{tr}(A)\operatorname{tr}(B).
\end{align}
This means either $a$ or $b$ must be zero. So, whatever conditions you have must rule out scalar multiples of the identity.
In general, let (you can do this by diagonalizing one of the matrices)
\begin{align}
A =
\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}
\ \ \text{ and } \ \ 
B =
\begin{bmatrix}
\lambda_1 & 0\\
0 & \lambda_2
\end{bmatrix}.
\end{align}
Then
\begin{align}
 \operatorname{tr}(AB)= \lambda_1 a+\lambda_2 d = (a+d)(\lambda_1+\lambda_2)=\operatorname{tr}(A)\operatorname{tr}(B)
\end{align}
if and only if $\lambda_1 d+\lambda_2 a = 0$. This means $[\lambda_1, \lambda_2]^T\perp [d, a]^T = M [a, d]^T$ where $M = \begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix}$ .
