Is there any inverse-commutator for matrices

Let $\operatorname{spectrum}(A)=(\lambda_i)_i$ and $f:A\rightarrow I\otimes A-A^T\otimes I$. Then $\operatorname{spectrum}(f)=\{\lambda_i-\lambda_j\mid i,j\}$. Assume that we are in the generic case, which implies that the $(\lambda_i)_i$ are pairwise distinct. Thus $\dim(\ker(f))=n$ and $\dim(\operatorname{im}(f))=n^2-n$. Thus there are solutions in $X$ if $C$ is in the orthogonal of $\ker(f)$ (in a sense to be defined); in particular, a necessary condition is $\operatorname{tr}(C)=0$.

Not sure if there is a closed-form answer, but $$ AX-XA = C $$ is a linear system of $n\times n$ equations of $n \times n$ unknowns.

One approach is using the vectorization operator $$\operatorname{vec} X = \begin{pmatrix} x_{11}\\ x_{21}\\ \vdots\\ x_{n1}\\ x_{12}\\ \vdots\\ x_{nn} \end{pmatrix}$$ Using Kronecker product $$ \operatorname{vec} (AXB) = (B^\top \otimes A) \operatorname{vec} X $$ the system becomes $$ (I \otimes A - A^\top \otimes I)\operatorname{vec} X= \operatorname{vec} C\\ \operatorname{vec} X = (I \otimes A - A^\top \otimes I)^{-1}\operatorname{vec} C \mkern{-227mu}\frac{\phantom{\operatorname{vec} X = (I \otimes A - A^\top \otimes I)^{-1}\operatorname{vec} C}}{\phantom{b}} $$

Edit. As kindly pointed by @loup blanc the matrix $I \otimes A - A^\top \otimes I$ is always degenerate so there is either no solution to the equation or infinite number of the solutions.

The another approach may be the following: let $\Omega \Lambda \Omega^{-1}$ be the eigendecomposition for $A$. $$ [A,X] = \Omega \Lambda \Omega^{-1} X - X \Omega \Lambda \Omega^{-1} = \Omega [\Lambda, \Omega^{-1}X\Omega] \Omega^{-1} = C\\ [\Lambda, \Omega^{-1}X\Omega] = \Omega^{-1}C\Omega $$ Denoting $R = \Omega^{-1}C\Omega, Y = \Omega^{-1}X\Omega$ the equation becomes $$ [\Lambda, Y] = R $$ with indices that is $$ \lambda_i Y_{ij} - \lambda_j Y_{ij} = R_{ij}\\ Y_{ij} = \begin{cases} \dfrac{1}{\lambda_i - \lambda_j}R_{ij}, &\lambda_i \neq \lambda_j\\ \text{any}, &\lambda_{i} = \lambda_j \end{cases} $$ provided that $R_{ij} = 0$ for every $\lambda_i = \lambda_j$ (at least, $\operatorname{diag}(R) = 0$).