I am studying the following hamiltonian which is very relevant for the electronic transport community, consisting of a double quantum dot system
\begin{align} \hat{H}_{2QD} = \sum_{i=1,2}v_{i}\hat{n}_{i}+\sum_{i=1,2}U_{i}\hat{n}_{i\uparrow}\hat{n}_{i\downarrow} + U_{12}\hat{n}_{i}\hat{n}_{j} + \hat{H}_{coupling} + \hat{H}_{leads} \end{align} where $\hat{n}_{i\sigma}=\hat{d}_{i\sigma}^{\dagger}\hat{d}_{i\sigma}$ is the number operator, $\hat{n}_i=\hat{n}_{i\uparrow}+\hat{n}_{i\downarrow}$ and $\hat{d}_{i\sigma}$ ($\hat{d}_{i\sigma}^\dagger$) are the annihilation (creation) operators for orbital $i$ and spin $\sigma$. Also, $v_{i}$ and $U_{i}$ are the on-site energy (also referred as gate level) and the inter-Coulomb repulsion of the level $i$ and $U_{12}$ is the intra-Coulomb repulsion between the levels 1 and 2. $\hat{H}_{leads}$ correspond to the single-particle eigenstates of the isolated leads. Finally $\hat{H}_{coupling}=\sum_{i=1,2}\sum_{k,\sigma}\sum_{\alpha=L,R}(t_{\alpha k}\hat{c}^{\dagger}_{\alpha k \sigma}\hat{d}_{i\sigma}+\text{H.c.})$ represent the coupling term between the quantum dots and the semi-infinite leads, where $\hat{c}_{\alpha\sigma}$ ($\hat{c}_{\alpha\sigma}^{\dagger}$) are the anihilation (creation) operators of the lead $\alpha$.
For practical reasons the coupling to the lead $\alpha$ is defined as $\gamma_{\alpha}=\sum_{k}\lvert t_{\alpha k }\rvert^2$.
This hamiltonian is known to be $SU(4)$ invariant when $\gamma_{1}=\gamma_{2}$ and $SU(2)$ invariant when $\gamma_{2}\to0$ (at $T\to 0 K$). I am wondering which is the transformation that shows explicitely this invariance.
I know that in the $SU(2)$ group a general rotation with Euler angles $\alpha,\beta,\gamma\in\Re$, can be written as
\begin{align} R(\alpha,\beta,\gamma) = e^{-iJ_3\alpha}e^{-iJ_2\beta}e^{-iJ_3\gamma}. \end{align}
I naively wonder if then $\hat{H}_{2QD}$ is said to be invariant under SU(2) rotations if $[\hat{H}_{2QD},R(\alpha,\beta,\gamma) ]=0\quad\forall \alpha,\beta,\gamma\in\Re$, or ,equivalently
\begin{align} \label{eq_H} \hat{H}_{2QD} = e^{iJ_3\alpha}e^{iJ_2\beta}e^{iJ_3\gamma}\hat{H}_{2QD}e^{-iJ_3\alpha}e^{-iJ_2\beta}e^{-iJ_3\gamma}, \end{align} where $J_i=\frac{\sigma_i}{2}$ being $\sigma_i$ the Pauli matrices.
