hope you are doing well.
Given a matrix $A$, I have to solve for $X$ (matrix) the equation $A^2X -2AXA +XA^2=0$, where both $A$ and $X$ are Hermitian. Looking for some related references, I found this one (see 5.1.10 and 5.1.11). They solve problems of the form $AX+XB=C$ (Lyapunov equation) and $\sum_n A_nXB_n=C$ (encapsulating sum), which are the terms contained in the original problem.
Thanks for your time devoted!
As @mathcounterexamples.net said, if you set the Endomorphism: $$\begin{array}{rrcl} F_A:& M_n(\mathbb{C}) &\to& M_n(\mathbb{C})\\ &X & \mapsto & AX-XA \end{array}$$ You have $F_A^2 (X) = A(AX-XA) - (AX-XA)A = A^2X - 2AXA + X A^2$, so your solutions are the kernel of $F_A^2$. We should study it a bit.
Since $A$ is Hermitian, there exists $U \in U_n(\mathbb{C})$ and $D = \operatorname{diag}(\lambda_1,\dots,\lambda_n)$ with $\lambda_i \in \mathbb{R}$ such that $$UAU^* = D$$ We thus only need to study $F_D$ since $(UF_AU^*)(U^*XU) = DX-XD = F_D(X)$.
With the canonical base $E_{i,j}$, we get $F_D(E_{i,j}) = (\lambda_i-\lambda_j) E_{i,j}$, so they are already all eigenvectors which means $F_D$ is already diagonal, how convenient!
Now we are looking for the eigenvectors associated to $0$ in $F_D^2$, which are exactly the eigenvectors of $F_D$ associated to $0$ since $F_D$ is diagonal which means that $$F_A^2(X)= 0 \iff F_D^2(UXU^*) = 0 \iff F_D(UXU^*) = 0 \iff F_A(X)=0$$ So your solutions are exactly the kernel of $F_A$ which is the commutant of $A$ and since $A$ is diagonalizable, if all its eigenvalues are different: $$S = \mathbb{C}[A].$$
