 Solution

Any constant real orthogonal matrix $R$ acting by $\mathbf{\varphi }\mapsto R\mathbf{\varphi }$ 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 $v\ne 0$, the vacuum manifold is the circle ${\mathbf{\varphi }}^T\mathbf{\varphi }=v^2$. Expand about the chosen point as ${\varphi }_1=v+h$, ${\varphi }_2=\pi$. The potential becomesThe kinetic terms remain canonically normalized real-scalar kinetic terms. Comparing the quadratic potential with $m_h^2{h}^2/2+m_{\pi }^2{\pi }^2/2$ givesThere 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.