 Solution

Both the canonical kinetic term and $V({\varphi }^{\mathsf{T}}\varphi )$ are invariant under the orthogonal group $O(n)$. The stated nonzero vacuum expectation value has length $|v|$. Its stabilizer is the orthogonal group $O(n-1)$ acting on the perpendicular subspace. HenceThe two nonzero coordinates of this particular vacuum do not break more generators than an axis-aligned vacuum: the two choices are related by a common orthogonal transformation.

Explicitly, with $s={\varphi }^{\mathsf{T}}\varphi$,stationary point at nonzero $v$ requires $V^{\prime }(v^2)=0$. The $n-1$ transverse eigenvalues vanish, while the radial squared mass is $4v^2{V}^{\prime \prime }(v^2)$. If this curvature also vanishes, there is an additional accidental massless radial mode; the number of symmetry-protected Goldstone bosons remains $n-1$. If $v=0$, the symmetry is unbroken and none is required.