 Solution

At $m_1=m_2$, the explicitly time-dependent coefficient becomes one. This restores time-translation symmetry, and the total energy becomes conserved. No further analogous restoration occurs merely from setting $m_3$ equal to them: the three arbitrary interaction functions still distinguish their arguments and particle labels.

If the question counts the extra ordinary unweighted rotations of the $x$ coordinate labels, the conventional generator count is as follows. For $m_1=m_2\ne m_3$, rotations of $(x_1,x_2)$ preserve both the kinetic and oscillator terms, so one ordinarily displayed rotation charge joins the restored energy: two displayed generators beyond the elementary list. For three equal masses, the full ordinary rotation group of $(x_1,x_2,x_3)$ has three generators; together with energy this gives four, or two more than in the preceding equal-pair case.

There is a necessary qualification to interpreting those numbers as genuinely new independent conservation laws. The weighted charges in solution deform continuously to those same ordinary rotation charges at equality. For example ${\mathcal{J}}_{12}$ is conserved for every positive $m_1,m_2$ and becomes $x_1{p}_{x_2}-x_2{p}_{x_1}$ when the masses agree. ThusThe restriction must be stated if using the first count. Equal mass does not create three previously absent independent oscillator integrals. If $\lambda =0$, energy conservation was already present; special choices of the interaction functions can add further symmetries and are excluded by the generic assumption.