Given a linear operator T on V over $\mathbb{R}$ is there a easy equivalent condition for the following
$$ \exists S : ST+TS = id_V $$
An easy necessary condition is $null(T) \leq T(V)$ and T invertible is sufficent but not necessary.
An idea I had was to extend a basis of $null(T)$ and work with matrices w.r.t this basis. But it's not clear whether this will yield a simple condition on T.
A more precise hypothesis which I suspect is false is that my necessary condition is sufficent, but I have not shown it to be false.
You should transform condition $ST+TS=I$ into a form involving the Kronecker product $\otimes$ i.e.,
$$(I \otimes T+T \otimes I)vec(S)=vec(I) \ \ (1)$$
isolating in this way the unknown $S$.
The Kronecker Product and some of its properties are defined in
https://en.wikipedia.org/wiki/Kronecker_product
Notation $vec(M)$ means the transformation of a matrix $M$ of any size $n \times p$ into the "piling" of all its columns of a single column vector with $n \times p$ elements.
The sufficient condition of existence given by @Omnomnomnom is a consequence of (1) because the eigenvalues of the $n^2 \times n^2$ matrix $M = I \otimes T+T \otimes I$ are all the $n^2$ possible sums $\lambda_i+\lambda_j$ where $\lambda_k$ designates any eigenvalue of $M$.
Edit (changed after a remark of @Davik; there was a confusion in my text between an "if and only if" and an "if" condition).
As a conclusion, system (1) ($n^2$ equations with $n^2$ unknowns) has a solution in two cases:
(guaranteed) if its kernel is $\{0\}$, i.e., if $0$ is not an eigenvalue of $I \otimes T+T \otimes I$, i.e., if $T$ has no opposite eigenvalues. In such a case, the solution is unique.
(contingent, including the first case) if $vec(I) \in Col(I \otimes T+T \otimes I)$. There may be in this case several solutions.