= Solution
Write $R^2=\sum_ix_i^2$ and impose tracelessness by a <Lagrange multiplier> $\zeta$. Differentiating $V-\zeta\sum_i x_i$ gives
$$
\boxed{4b x_i^3+(4aR^2-2m^2)x_i-\zeta=0\qquad(i=1,\ldots,5).}
$$
For $b\ne0$, each <eigenvalue> is a real root of the same nonzero cubic <polynomial>, so at most three distinct values occur at any stationary point, in particular at any existing minimum. If the multiplicities are $n_1,\ldots,n_k$, the unbroken subgroup is the block centralizer
$$
\boxed{S\bigl(U(n_1)\times\cdots\times U(n_k)\bigr),\qquad
k\le3,\quad\sum_\alpha n_\alpha=5.}
$$
Its connected Lie algebra is $\bigoplus_\alpha\mathfrak{su}(n_\alpha)\oplus\mathfrak u(1)^{k-1}$. The partitions allowed by this upper bound are $5$, $4+1$, $3+2$, $3+1+1$, and $2+2+1$. These are candidate patterns, not a proof that every one is a stable minimum for given couplings. For three distinct roots, the absent quadratic coefficient also implies that their unweighted sum is zero; tracelessness imposes an additional multiplicity-weighted relation.
\b[The unrestricted printed claim has an exception.] If $b=0$, $a>0$ and $m>0$, every traceless matrix with $R^2=m^2/(2a)$ is a global minimum. For example
$$
\Phi_d=\frac{m}{\sqrt{20a}}\operatorname{diag}(-2,-1,0,1,2)
$$
has five distinct <eigenvalues> and attains that minimum. Its unbroken subgroup is the maximal torus $U(1)^4$. The cubic argument therefore requires a nonzero $b$, or another condition excluding the identically vanishing stationarity polynomial. Couplings for which the potential is unbounded below have no global minimum to classify.
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