About generator of $SU(2)$ flavor symmetry group I am reading the textbook "Weak Interactions" by Howard Georgi which can be found in his homepage.
Here, I am trying to solve problem 1b-2. The problem is given as follows. 


Consider the Lagrangian 
  \begin{align}
i \bar{\psi}_1 {D\!\!\!/}_{\mu}   \psi_1+i \bar{\psi}_2 {D\!\!\!/}_{\mu}   \psi_2
\end{align}
  Where $D^{\mu}_j = \partial^{\mu}_j - iG_a^{\mu} T_j^a$, 
  with $T_1^a = \frac{1}{2}\tau_a$
  where $\tau_a$ are the Pauli matrices but 
  \begin{align}
T_2^1 = \frac{1}{2}\tau_2, \quad T_2^2 = \frac{1}{2}\tau_3, \quad T_2^3 = \frac{1}{2}\tau_1
\end{align}
  This theory has an $SU(2)$ gauge symmetry, and also an $SU(2)$ global 
  flavor symmetry, because $\psi_1$ and $\psi_2$ transform under equivalent representations of the gauge symmetry which we have chosen, perversely, to write in different forms. Find the generators of the $SU(2)$ flavor symmetry. 


First Guess:
Of course the flavor symmetry has $SU(2)$, thus its generators are written in terms of Pauli matrices, $\tau$. 
But from the fact that generators of flavor symmetry and gauge symmetry  commute, I got confused.  
Of course, I can think that the group actions for flavor symmetry and gauge symmetry are different (Can think of direct product, $i.e$, $SU(2)_f \times SU(2)_g$), so commuting of two different action is naturally guaranteed. 
(The above problem states: find the generators of the $SU(2)$ flavor symmetry;  so applying the above logic, it is natural to pick up $\tau$, I guess.)
I think this is not the author's intention. 
Can anyone give some explicit explanation or computation on this problem?
 A: Elapsed time ensures this is not doing anyone's homework. The problem itself is open-ended and I gather the technical second part of the answer below is not needed/called for, but this is the more instructive and technically sweet part of the problem students of formal rotations might appreciate. The essence of the problem is to appreciate that there is entwining between the gauge group transformations and interchanges of the two "flavors",  so you must unravel the entwining such that gauge transformations act completely identically on either fermion, so gauge and flavor commute, as per the Cartesian product structure your thought about.
Here, the group action is strictly different for each fermion 1 or 2; but, since the Lie algebra of SU(2) is isomorphic to cyclic permutations of its generators, the representations for fermions 1 and 2 are actually equivalent, that is 
$$
(\tau_1,\tau_2,\tau_3) = U   (\tau_3,\tau_1,\tau_2) U^\dagger
$$
for some unitary 2×2 matrix (possibly not required to be found explicitly— but this is the  tangential part deferred to the end).
That is to say, the "perverse" representation for the second fermion is a cyclic permutant of the "natural" one for the first,
$$
T_2^a =U T_1^a U^\dagger.
$$
The corresponding coupling terms then reduce to 
$$
\bar{\psi}_1 { G \!\!\!/} _a  T_1^a   \psi_1+  \bar{\psi}_2  { G \!\!\!/} _a U T_1^a U^\dagger  \psi_2 .
$$
It is then evident both components of the doublet $\tilde \psi_j\equiv (\psi_1, U^\dagger \psi_2)$ transform identically under the gauge SU(2), through $T_1^a$, which is therefore in a separate Cartesian product factor from the global flavor SU(2). 
The flavor SU(2) mixes up the two components of the isodoublet $\tilde \psi$, indexed by j through SU(2)-doublet matrices $\tau_f^{ij}$, which are completely impervious to the gauge generators coupling to G.
The above two coupling terms then, like the kinetic terms, constitute flavor SU(2) singlets! 
My sense is that this is what the problem asked the routine student to appreciate.  Still ...

How to find U?       This is the technically sweet and instructive rotation point worth an answer—but probably merely tangential to the problem:  What is the equivalence transformation permuting the Pauli matrices cyclically?
The Pauli vector rotates to 
$$
\vec \tau \mapsto   \begin{pmatrix}
      0               &   0& 1\\
        1               &   0& 0\\
 0               &   1& 0
    \end{pmatrix} ~  \vec \tau   =\exp \left ( \frac{2\pi}{3}  \begin{pmatrix}
      0               &   -1& 1\\
        1               &   0& -1\\
            -1              &   1& 0
    \end{pmatrix}/\sqrt{3} \right  ) ~~~  \vec \tau 
$$
First,  visualize the coordinate axes in 3-space and observe that a cyclic permutation of the three axes, x,y,z is achievable through a $2\pi/3$ rotation around the fully symmetric axis $\hat n=(1,1,1)/\sqrt{3}$ in the principal octant. The above expresses this through Rodrigues' rotation formula.
The adjoint indices of the Pauli matrices then also transform under the bifundamental action by 
$$
\tau_a \mapsto ( -1\!\! 1 /2+i\hat n \cdot \vec \tau /2  )^\dagger  \tau_a  (-1\!\! 1 /2+i\hat n \cdot \vec \tau \sqrt{3}/2  ),
$$
where the cosine, -1/2, and sine, $\sqrt{3}$/2,  of $2\pi/3$ have been plugged in the finite rotation formula, $U=\exp (i\frac{2\pi}{3}\hat n \cdot \vec \tau )$. (See, e.g., this answer if not obvious.)
Working out the explicit form of the equivalence matrices amounts to 
$$
U= (-1\!\! 1+i(\tau_1+\tau_2+\tau_3 ))/2= \frac{1}{2}    \begin{pmatrix}
      i-1                &   i+1\\
      i-1 &  -1-i
    \end{pmatrix}   , \qquad U^\dagger = - \frac{1}{2}    \begin{pmatrix}
      i+1                &   i+1\\
      i-1 &  1-i
    \end{pmatrix} ,
$$
which checks to effect the cyclic permutation desideratum.
