Let $\mathfrak{g}$ be a complex simple Lie algebra with bracket $[x,y]$. For **which** $z\in \mathfrak{g}$ does the formula
$$
\mu(x,y)=ad (z)([x,y])=[z,[x,y]]
$$
define another Lie bracket on the same vector space ? For $\mathfrak{g}=\mathfrak{sl}(2,\mathbb{C})$ this holds for every $z$. Explicit computation seems to show that for $\mathfrak{g}=\mathfrak{sl}(n,\mathbb{C})$ with $n\ge 3$ only $z=0$ is possible. Does this hold for all simple Lie algebras of rank at least $2$, and is there a proof somewhere ?   

In the literature, often the Jacobi identity for $\mu(x,y)=[z,[x,y]]$ is studied for **all** $z$. Then it is known that only $\mathfrak{sl}(2,\mathbb{C})$ is possible. Another reference are simple Hom-Lie algebras, but this is different.
For nonassociative algebras a new multiplication $x\circ\_{z} y$ on the same vector space induced by an element $z$ is called a *homotope*.


What is known for simple Lie algebras ?


EdiT: I saw that $[x,y]\_R:=[R(x),y]+[x,R(y)]$ is a Lie bracket iff $R$ is a classical $R$-matrix. For $R=ad(z)$ we have $[x,y]\_R=[z,[x,y]]$. So the question is, for which $z$ the map $R=ad(z)$ is a classical $R$-matrix of a simple Lie algebra.



Homotopes of simple Lie algebras
In the paper [Derivation Double Lie Algebras](https://homepage.univie.ac.at/Dietrich.Burde/papers/burde_49_double_lie.pdf) the following result is proved:


*Theorem $3.2$*: Let $\mathfrak{g}$ be a simple Lie algebra of rank $r\ge 2$ over an algebraically closed field $K$ of characteristic
zero, and $z\in \mathfrak{g}$. Suppose that $R={\rm ad}(z)$ is a classical $R$-matrix. Then $z=0$ and $R=0$.


The proof uses the Jordan-Chevalley decomposition for $z=n+s$, the Jacobson-Morozov theorem, arguments concerning roots and algebraic group actions etc. This gives $s=0$ and ${\rm ad}(z)^3=0$, and finally $z=0$.
For the case of rank $1$ we have the following result:


*Proposition $3.1$*: Let $\mathfrak{g}=\mathfrak{sl}(2,\mathbb{C})$. Then $R={\rm ad}(z)$ is a classical $R$-matrix for all $z\in \mathfrak{g}$.
The Lie algebra $\mathfrak{g}\_R$ is isomorphic to $\mathfrak{r}\_{3,1}(\mathbb{C})$ for all $z\neq 0$.


Here $\mathfrak{r}\_{3,1}(\mathbb{C})$ is the $3$-dimensional solvable Lie algebra given by $[e\_1,e\_2]=e\_2$ and $[e\_1,e\_3]=e\_3$, and $\mathfrak{g}\_R$ has bracket $[x,y]\_R=[z,[x,y]]$.


