Let $A$ be semipositive stable, i.e., real part of the eigenvalues of $A$ are nonnegative. Let $L_{A}: S^{n} \rightarrow S^{n}$ be a linear transformation defined by $L_{A}(X)=AX+XA^T$. I want to construct an example such that $A$ is semipositive stable but group inverse of $L_{A}$ does not exit, i.e., $L_{A}$ is not group invertible.
Note: Group inverse of a matrix $B \in \mathbb{R}^{n\times n}$ if exists is the unique matrix $X$ such that $AXA=A$, $XAX=X$ and $AX=XA$. Group inverse of $B$ is denoted by $B^{\#}$.
Example: Here I give one example where $A$ is semipositive stable and $L_{A}^{\#}$ (group inverse of $L_{A}$) exists. $A=\begin{pmatrix} 1 & 1\\ 0 &0 \end{pmatrix}$. Set $X=\begin{pmatrix} x_{11} & x_{12}\\ x_{12} & x_{22} \end{pmatrix}$. Then $L_{A}(X)=\begin{pmatrix} 2(x_{11}+x_{12}) & x_{12}\\ x_{12} & x_{22} \end{pmatrix}$ and $L_{A}^{\#}(X)=\begin{pmatrix}\frac{1}{2}(x_{11}+x_{12}) & x_{12}\\ x_{12} & x_{22} \end{pmatrix}$ (one can verify all the three conditions of group inverse for $L_{A}$).
Try $A = 0$.
For a $2 \times 2$ matrix $A$, $L_A$ is non-invertible if $A$ is non-invertible or $\text{Tr}(A) = 0$.
EDIT: OK, try $$A = \pmatrix{0 & 1\cr 0 & 0\cr}$$
BTW, where does this "group inverse" concept come from? It's a strange name, as it doesn't seem to have to do with groups.