Invariance of a symmetric sum

Consider two matrices $$M$$ and $$N$$ of same dimension, such that $$L=MN+NM$$ is the symmetric sum in the sense that interchanging $$M$$ and $$N$$ does not change $$L$$. Is there any transformation $$T[M]=M^\prime$$ and $$T[N]=N^\prime$$ such that $$M^\prime N^\prime+N^\prime M^\prime=MN+NM$$? In other words, the symmetric sum is invariant under the transformation $$T$$.

• Yes, the identity. Surely you want something else? Commented Oct 7, 2022 at 20:00

Not a full answer, but have you considered looking at what happens if the dimension is low? Suppose $$M,N\in\mathbb{R}^\ast=\text{GL}_1(\mathbb{R})$$ (we don't need to consider $$0$$, otherwise $$L=0$$). Then we have $$MN+NM=2MN$$ by commutativity.
Suppose $$T$$ is a linear transformation from $$\mathbb{R}$$ to itself. Then, your condition translates to $$2MN=T(M)T(N)$$. Since $$T$$ is linear, we have $$T(M)T(N)=MT(1)NT(1)=MNT(1)^2$$. Thus, $$L=2MNT(1)^2$$.
Since $$M,N\in\mathbb{R}^\ast$$, we can divide by $$2MN$$ on both sides to get $$T(1)^2 = 1$$, so $$T(1) = \pm 1$$, which means $$T(x)=\pm x$$ works. This already gives you two possible transformations, $$\pm \text{Id}_n$$. Using a basis (assuming your vector spaces are finite), I think those are the only two possible such transformations but you'd need to check by hand, which seems a bit tedious. Maybe someone with a better idea comes along, but at least this gives you two basic transformations that do what you want.
Suppose the elements of the matrices are taken from a field whose characteristic is not $$2$$. Then the only possible choices of $$T$$ are $$\pm\operatorname{Id}$$.
Let $$A=T(I)$$. The given condition implies that $$T(X)A+A\,T(X)=2X\tag{1}$$ for every matrix $$X$$. In particular, if we put $$X=I$$, we obtain $$A^2=I$$. Hence $$A$$ is diagonalisable and its only possible eigenvalues are $$1$$ and $$-1$$.
Suppose both $$1$$ and $$-1$$ are eigenvalues of $$A$$. Then $$v^TA=-v$$ and $$Au=u$$ for some nonzero vectors $$u$$ and $$v$$. But then we may pick two vectors $$x$$ and $$y$$ such that $$v^Tx=y^Tu=1$$ and with $$X=xy^T$$, $$(1)$$ implies that $$2=2v^T(xy^T)u=v^T\left(T(X)A+A\,T(X)\right)u =v^TT(X)(Au)+(v^TA)T(X)u=0,$$ which is a contradiction. Therefore $$A=\pm I$$ and $$(1)$$ implies that either $$T(X)=X$$ for all $$X$$ or $$T(X)=-X$$ for all $$X$$.