The matrix equation $L^* \rho L+ L \rho L^* = L^* L \rho + \rho L^* L$ I want to find all $n \times n$ matrices $L$ that satisfy the following equation
\begin{align*}
L^* \rho L+ L \rho L^* = L^* L \rho + \rho L^* L
\end{align*}
for all selfadjoint $n \times n$ matrices $\rho$. Here $L^*$ denotes the adjoint of $L$. 
 A: I assume that the matrices are real, $L^*=L^T$ and that $\rho$ is denoted by $R$ s.t. $R=R^T$.
$\textbf{Proposition}$. The solutions in $L$ are only all the scalar matrices.
$\textbf{Proof}$. Step 1. Let $(E_{i,j})_{i,j}$ be the canonical basis of $M_n(\mathbb{R})$. Here, for every $i$, 
$L^TE_{i,i}L+LE_{i,i}L^T-L^TLE_{i,i}-E_{i,i}L^TL=0$. Seing the diagonal elements, we obtain that (except the entry $(i,i)$), the row $i$ and the column $i$ of $L$ are $0$; finally $L=diag((l_i)_i)$ is diagonal.
Step 2. Now the function $f:R\in S_n\mapsto 2LRL-L^2R-RL^2$ is $0$, that is,
$f=2L\otimes L-L^2\otimes I-I\otimes L^2=0$ over $S_n\subset \mathbb{R}^{n^2}$, where we stack the matrices into vectors, row by row.
cf. https://en.wikipedia.org/wiki/Kronecker_product
From $f=-(L\otimes I-I\otimes L)^2$, we deduce that 
$spectrum(f)=(-(l_i-l_j)^2)_{i,j}$ and $f$ is diagonalizable.
It is not difficult to see that $f=0$ over $\mathbb{R}^{n^2}$ and, consequently, the $(l_i)_i$ are equal and $L$ is a scalar matrix.
EDIT. For Step 2, if you don't like the tensor products, then you can proceed as follows
When $i\not= j$, $f(E_{i,j})=-(l_i-l_j)^2E_{i,j},f(E_{j,i})=-(l_i-l_j)^2E_{j,i}$. Since $f(E_{i,j}+E_{j,i})=0$, we deduce that $l_i=l_j$.
