= Solution
Represent the <color charges> by $r=(1,0,0)^T$, $g=(0,1,0)^T$ and $b=(0,0,1)^T$. The <Gell-Mann matrices> satisfy
$$
\lambda^4r=b,\qquad\lambda^5r=ib,\qquad
\frac{\lambda^4-i\lambda^5}{2}=|b\rangle\langle r|.
$$
The physical <quark>-<gluon> <Feynman vertex> contains $g_sT^a$ with $T^a=\lambda^a/2$; the off-diagonal combination therefore allows $r\to b$. <Color charge> conservation assigns the emitted <gluon> the colour-flow label $r\bar b$.
\b[The claimed impossibility of $r\to r$ is false.] A direct counterexample from the printed matrices is
$$
\boxed{\lambda^3r=r,\qquad\lambda^8r=\frac{r}{\sqrt3}.}
$$
Thus both diagonal <gluons> have nonzero $r\to r$ vertex elements, $(T^3)_{rr}=1/2$ and $(T^8)_{rr}=1/(2\sqrt3)$. They correspond to the traceless combinations
$$
\frac{r\bar r-g\bar g}{\sqrt2},\qquad
\frac{r\bar r+g\bar g-2b\bar b}{\sqrt6}.
$$
A <Gell-Mann matrix> is a generator rather than a finite rotation, but exponentiating a diagonal generator also leaves the red basis direction fixed up to a phase. The actual excluded object is the identity generator, proportional to the <colour singlet> $(r\bar r+g\bar g+b\bar b)/\sqrt3$. Adding it would extend <SU(3)> to <unitary group> and introduce a ninth singlet vector boson. A colour-preserving vertex among the existing eight <gluons> does \b[not] require that extra boson and does not make the ordinary diagonal <gluons> colour singlets.
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