= Solution
Any constant real <orthogonal matrix> $R$ acting by $\boldsymbol\phi\mapsto R\boldsymbol\phi$ preserves both the Euclidean field-space norm and the kinetic term. The full internal symmetry is $O(2)$, with connected component $SO(2)\simeq U(1)$; reflections add a discrete component.
For $v\ne0$, the vacuum manifold is the circle $\boldsymbol\phi^T\boldsymbol\phi=v^2$. Expand about the chosen point as $\phi_1=v+h$, $\phi_2=\pi$. The potential becomes
$$
V=4\lambda v^2h^2+4\lambda v h(h^2+\pi^2)+\lambda(h^2+\pi^2)^2.
$$
The kinetic terms remain canonically normalized real-scalar kinetic terms. Comparing the quadratic potential with $m_h^2h^2/2+m_\pi^2\pi^2/2$ gives
$$
\boxed{m_h^2=8\lambda v^2,\qquad m_\pi^2=0.}
$$
There is one massive radial mode and one massless tangential <Goldstone boson>. A chosen vacuum leaves only the reflection that fixes that point, so there is one broken continuous generator. For negative $v$ the chosen point lies on the opposite side of the same circle and the mass depends on $v^2$; the radius is $|v|$. If $v=0$, the vacuum circle collapses to the origin, no continuous generator is broken, and both quadratic masses vanish without either field being a <Goldstone mode>. The nonzero-$v$ assumption is essential to the spontaneous-breaking interpretation.
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