= Solution
Let $V(\phi)=m^2\phi^2/2+\lambda(\phi^2)^2/4$, where $\phi^2=\sum_i\phi_i^2$. Its stationary-point equation is $(m^2+\lambda\phi^2)\phi_i=0$.
For $m^2>0$, the unique minimum is $\phi=0$. The quadratic <mass term> is $m^2\delta_{ij}$, so \b[there are $n$ <real scalar fields>, all of mass $\sqrt{m^2}$], and the global <special orthogonal group> $SO(n)$ remains unbroken.
For $m^2<0$, write $v^2=-m^2/\lambda$. The <vacuum manifold> is the sphere $\phi^2=v^2$. Choose $\langle\phi\rangle=v e_n$. The transformations preserving this <vacuum expectation value> rotate the first $n-1$ components, giving \b[$SO(n)\to SO(n-1)$]. At this vacuum the potential's <Hessian matrix> is
$$
\left.\frac{\partial^2V}{\partial\phi_i\partial\phi_j}\right|_{ve_n}
=2\lambda v^2\delta_{in}\delta_{jn}.
$$
Writing $\phi=(\pi_1,\ldots,\pi_{n-1},v+\eta)$, the radial mode $\eta$ therefore has
$$
\boxed{m_\eta^2=2\lambda v^2=-2m^2,\qquad m_{\pi_a}^2=0\quad(a=1,\ldots,n-1).}
$$
These are one massive radial scalar and $n-1$ physical <Goldstone bosons>. The <Goldstone theorem> applies to the broken global generators; the flat angular directions of the <vacuum manifold> explain their zero masses. The <spontaneous symmetry breaking> is a choice of vacuum, not a change in the invariant Lagrangian. For $n=1$ there are no continuous angular directions: only the radial scalar remains, and the potential's discrete $\phi\mapsto-\phi$ symmetry is broken.
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