Do two Lie brackets differ by a matrix? I'm thinking about the following seemingly simple problem. Let $V = \mathbb{R}^n$, and $[\,\cdot\,, \cdot\,]_1$,
$[\,\cdot\,, \cdot\,]_2$ be two distinct Lie brackets on $V$.

Is there always some $A \in \operatorname{End}(V)$ such that either
$A [\,\cdot\,, \cdot\,]_1 = [\,\cdot\,, \cdot\,]_2$ or
$A [\,\cdot\,, \cdot\,]_2 = [\,\cdot\,, \cdot\,]_1$?

For example, if $[\,\cdot\,, \cdot\,]_1$ is arbitrary and
$[\,\cdot\,, \cdot\,]_2$ is trivial (always $0$), then we can simply take the $0$ matrix $A$ and obtain the result. Now of course if the two brackets are swapped in position than this no longer holds, so the order matters.
This seems natural to me, but I do not know how to approach this problem, other than trying to find counterexamples with low dimensional Lie algebras. But I am struggling there too. Any help would be appreciated, and if there is already known results about this (a proof or a counterexample) I would greatly appreciate a reference as well.
 A: As discussed in comments, the question seems a bit unclear. The way I interpret it is that (with your notation) you are asking if there is $A \in \mathrm{End}(V)$ such that at least one of either $A([x,y]_1)=[x,y]_2$, or $A([x,y]_2)=[x,y]_1$, holds for all $x,y \in V$.
No, as per the following counterexample:
On $\mathbb R^3$ with standard basis $e_1,e_2, e_3$, define a Lie bracket $[ \cdot, \cdot]_1$ by $[e_1,e_2]_1 = -[e_2,e_1]_1 = e_1$, all other $[e_i, e_j]_1 =0$. Define a second Lie bracket via $[e_2,e_3]_2= - [e_3,e_2]_2 = e_2$, all other $[e_i, e_j]_2=0$.
There cannot be an $A$ such that $A([e_2, e_3]_1) = [e_2,e_3]_2$ because the LHS is $0$ and the RHS is not. But neither can there be an $A$ such that $A([e_1,e_2]_2)=[e_1, e_2]_1$ because the LHS is $0$ and the RHS is not.
Note that this is the case even though the two Lie algebras are actually isomorphic; they both are of the form $\mathbb R \oplus S$ where $\mathbb R$ is the one-dimensional abelian and $S$ is the two-dimensional non-abelian Lie algebra.

Here is how I came up with this counterexample. The question can be reformulated as: If $f: V \otimes V \rightarrow V$ is the linear map induced by the first Lie bracket, and $g: V\otimes V \rightarrow V$ the one induced by the second Lie bracket, is there a linear map $A : V \rightarrow V$ in either direction which makes the obvious triangular diagram commute? And it's clear there cannot be one as soon as the kernels of $f$ and $g$ are not contained in one another, one way or the other. Admittedly, since $f$ and $g$, coming from Lie brackets, cannot just be any linear maps, something like this cannot be constructed for $\dim(V) \le 2$, but for $\dim(V)=3$, the above example offered itself.
