Rozansky-Witten invariants of hyperkahler manifolds and independence of complex structure

Recently I have been learning about Rozansky-Witten invariants, mainly through Hitchin-Sawon's paper "curvature and characteristic numbers of hyperkahler manifolds" and through Justin Sawon's phD thesis. I wanted to ask for explanations about a specific point in these references, namely the independence of choice of complex structure compatible with the hyperkahler structure. What follows is a more detailed version of this question; any help would be much appreciated!

Let $$(X,g,I,J,K)$$ be a compact hyperkahler manifold of real dimension $$4n$$. We make a choice and use $$I$$ to decompose the complexified tangent bundle $$TX\otimes_\mathbb{R}\mathbb{C}$$ as $$T\oplus \overline{T}$$, and similarly for complex differential forms on $$X$$. Fixing notation, we write $$\omega_I,\omega_J,\omega_K$$ the Kahler forms associated to $$I,J,K$$ and $$\omega = \omega_J + i\omega_K\in \Omega^{2,0}(X)$$, which is a holomorphic symplectic form. The curvature tensor can be written as $$K\in\Omega^{1,1}(X,\mathrm{End}T)$$, or using $$\omega$$ to identify the holomorphic tangent bundle $$T$$ with the holomorphic cotangent bundle $$T'$$ we can write it as $$\Phi\in\Omega^{0,1}(X,\mathrm{Sym}^3T')$$.

For any trivalent graph $$\Gamma$$ with $$2n$$ vertices, we define a $$(0,2n)$$ differential form $$\Gamma(\Phi)$$ by contracting the tensor $$\Phi^{\otimes 2n}\otimes \tilde{\omega}^{\otimes 3n}$$ along the edges of $$\Gamma$$, where $$\tilde{\omega}\in T\wedge T$$ is the dual form to $$\omega$$, and then anti-symmetrizing to obtain a differential form. See the references for details.The Rozansky-Witten invariant associated with $$\Gamma$$ and $$X$$ is then defined as $$b_{\Gamma}(X) = \frac{1}{(8\pi^2)^n n!}\int_X \Gamma(\Phi)\wedge\omega^{n}.$$ My question is the following: why doesn't $$b_{\Gamma}(X)$$ depend on the choice of $$(I,J,K)$$? More precisely, we could have chosen $$(I',J',K')$$ related to $$(I,J,K)$$ by any rotation in $$SO(3)$$ and the claim is that the invariant above stays the same.

Here is how I have interpreted the explanation given in Hitchin-Sawon and other papers by Sawon such as "Generalisations of Rozansky-Witten invariants" with Justin Roberts. The hyperkahler structure on $$X$$ can be expressed as $$TX\otimes_\mathbb{R}\mathbb{C} \approx E_{n}\otimes T$$ where $$E_n$$ is a complex vector bundle over $$X$$ of rank $$2n$$ endowed with non-degenerate form $$\epsilon\in\Lambda^2E_n^*$$ and $$T$$ is a trivialisable complex vector bundle of rank $$2$$ - some explanations about this would be very helpful, but mostly what I'm interested in is how to use this to prove the following claim.

Write the curvature tensor as $$\Omega\in\mathrm{Sym}^4E_n$$ using the isomorphism $$TX\otimes_\mathbb{R}\mathbb{C} \approx E_{n}\otimes T$$. We may use $$(\Omega,\epsilon)$$ as weights for our graphs instead of $$(\Phi,\tilde\omega)$$, so proceeding by before we associate to any $$\Gamma$$ trivialent graph of $$2n$$ vertices a form $$\Gamma(\Omega)\in\Lambda^{2n}E_n$$.Denoting by $$\tilde{\epsilon}\in\Lambda^2 E_n$$ the dual form of $$\epsilon$$, we have $$\Gamma(\Omega)=\Gamma_\epsilon(\Omega)\tilde{\epsilon}^n$$ for some function $$\Gamma_\epsilon(\Omega)$$ and then, denoting by $$d\mathrm{vol}$$ the volume form of $$(X,g)$$, $$b_\Gamma(X) = \frac{1}{(8\pi^2)^n n!}\int_X \Gamma_\epsilon(\Omega) d\mathrm{vol}$$ The claim is that the right-hand side is equal to the left-hand side, which is defined as before by making a choice of $$(I,J,K)$$. Since the right-hand side does not depend on this choice, we can deduce that $$b_\Gamma(X)$$ depends only on the hyperkahler structure and not on arbitrary choices of the generators $$I,J,K$$. How to prove this claim?

From the hypercomplex structure, any hyperkahler manifold $$X$$ admits an action of $$SU(2)$$ on $$TX$$, which naturally extends to the complexification $$TX\otimes_\mathbb{R}\mathbb{C}$$ and to all tensors. Let $$T=T(\sigma)$$ be a tensor obtained after fixing generators of the quaternionic structure $$\sigma=(I,J,K)$$. We say $$T$$ is $$SU(2)$$-equivariant if $$(U\cdot T)(\sigma) = T(U\cdot\sigma)$$, where $$U\cdot\sigma=(UIU^*,UJU^*,UKU^*)$$ and $$U\cdot T$$ denotes the action of $$U$$ on the tensor $$T$$. The statement is proved once we show that $$\Phi$$, $$\omega$$ and $$\tilde{\omega}$$ are $$SU(2)$$-equivariant in the sense above.
Indeed, assuming this then for any oriented trivalent graph $$\Gamma$$ with $$2n$$ vertices, where $$\dim_\mathbb{R}(X)=4n$$, we have that $$\Gamma(\Phi)\wedge\omega^n$$ is $$SU(2)$$-equivariant. Since it is a volume form, and the $$SU(2)$$ action is trivial on volume forms, we conclude that $$\Gamma(\Phi)\wedge\omega^n$$ is $$SU(2)$$-invariant - therefore the Rozansky-Witten invariants does not depend on the choice of $$\sigma$$. I am left with the exercise of proving that $$\Phi$$, $$\omega$$ and $$\tilde{\omega}$$ are $$SU(2)$$-equivariant.
For example, $$SU(2)$$-equivariance of $$\omega$$ seems to follow easily. Let $$\omega_L(u,v) = g(u,Lv)$$ for any complex structure $$L=aI+bJ+cK$$, parameterized by $$\mathbb{S}^2$$. Then: $$L = UIU^*\Rightarrow \omega_L(u,v) = \omega_{I}(U^* u , U^*v),$$ which shows $$SU(2)$$-equivariance since the right-hand side is $$(U\cdot\omega_I)(u,v)$$. The relation $$\omega_L(u,v) = \omega_{I}(U^* u , U^*v)$$ is straightforward to check: $$\omega_L(u,v) = g(u,Lv) = g(u,UIU^*v) = g(U^*u,IU^*v) = \omega_I(U^*u,U^*v).$$ This implies that $$\omega=\omega_J + i\omega_K$$ is $$SU(2)$$-equivariant; the equivariance of $$\tilde{\omega}$$ follows. Similarly, the equivariance of $$\Phi$$ follows from the invariance of the curvature tensor of a hyperkahler manifold, and this seems to answer my question completely!