In Huybrecht's book on complex geometry, he states the following lemma on page 193:
Lemma 4.4.2: The $k$-multilinear symmetric map $P$ is invariant if and only if for all $B,B_1,\ldots, B_k \in \mathfrak{gl}(r;\mathbb{C})$ one has
$$ \sum_{j=0}^k P(B_1, \ldots, B_{j-1},[B,B_j],B_{j+1},\ldots,B_k) = 0 $$
I'm unsure about the tricks used in the proof for both directions.
For the 'only if' direction, he gives the following hint: Use the invertible matrix $C_t = e^{tB}$ and differentiate the invariance equation $$ P(C_tB_1C_t^{-1},\ldots,C_tB_kC_t^{-1}) = P(B_1, \ldots,B_k) $$ and evaluate at $t=0$.
I first made a guess at what this would look like for a $1$-multilinear map:
$$ \begin{align*} \frac{d}{dt}P(C_tB_1C_t^{-1})|_{t=0} &= P(BC_tB_1C_t^{-1} - C_tB_1BC_t^{-1})|_{t=0} \\ &= P(BB_1 - B_1B) \\ &= P([B,B_1]) \\ &= \frac{d}{dt}P(B_1) = 0 \end{align*} $$
So as a naive guess, I went onto the general case by following the same procedure and looking at
$$ P([B,B_1], \ldots, [B,B_k]) = 0 $$
which has lead me nowhere. I attempted splitting up the polynomials from their brackets; e.g. setting $$ \begin{align*} P([B,B_1],[B,B_2]) &= P(BB_1,[B,B_2]) - P(B_1B,[B,B_2]) \\ &= P(BB_1,BB_2) - P(BB_1,B_2B) - P(B_1B,BB_2) + P(B_1B,B_2B) \end{align*} $$
But this does not seem fruitful.
In the reverse direction, I rearranged the equality getting $$ \sum_{j=1}^kP(B_1, \ldots,B_{j-1},BB_j,B_{j+1}, \ldots, B_k) = \sum_{j=1}^kP(B_1, \ldots,B_{j-1},B_jB,B_{j+1}, \ldots, B_k) $$
but I don't see how this could imply the desired result.
Denote $B\cdot P$ the action of an invertible matrix $B$ on $P$.
Hint. The group $GL_n(\mathbb C)$ is generated by exponentials, so to show that a map $P$ is invariant it is enough to show that it is invariant by matrices of the form $\exp(B)$. Now consider the map $t\mapsto \exp(tB)\cdot P$. It satisfies a certain first order differential equation, and therefore you can use the uniqueness theorem for ODEs to show that it vanishes identically from knowledge about what happens at just one point (the initial condition) Use that.