= Solution
The unbroken <group> is the <stabilizer subgroup> of the <vacuum expectation value>. For a nonzero vector <vacuum expectation value>, an <orthogonal transformation> puts $\langle\phi\rangle=v e_N$. A special orthogonal transformation fixing it has the block form $\operatorname{diag}(A,1)$ with $A\in\operatorname{SO}(N-1)$. Consequently
$$
\boxed{\operatorname{SO}(N)\longrightarrow\operatorname{SO}(N-1)\quad(v\ne0).}
$$
The zero <vacuum expectation value> leaves the entire <special orthogonal group> unbroken.
For a real <symmetric tensor> write its <vacuum expectation value> as a real <symmetric matrix> $Q$. The transformation is $Q\mapsto OQO^T$. The <spectral theorem for real symmetric matrices> gives an orthonormal <basis> in which
$$
Q=\operatorname{diag}(q_1I_{m_1},\ldots,q_pI_{m_p}),\qquad q_i\ne q_j\ (i\ne j),\qquad \sum_i m_i=N.
$$
Because $OQO^T=Q$ is equivalent to $OQ=QO$, an unbroken transformation preserves each <eigenspace>. The complete unbroken <group> is therefore
$$
\boxed{H=S\bigl(O(m_1)\times\cdots\times O(m_p)\bigr)
=\{(A_1,\ldots,A_p):\prod_i\det A_i=1\}.}
$$
Its <identity component> is $H^0=\prod_i\operatorname{SO}(m_i)$. This <symmetric tensor vacuum stabilizer> shows why the <representation> alone does not determine a unique breaking pattern:
* If $Q=qI$, all of $\operatorname{SO}(N)$ survives, even when $q\ne0$.
* If all <eigenvalues> are distinct, $H\cong(\mathbb Z/2\mathbb Z)^{N-1}$, consisting of diagonal sign changes with total sign $+1$; no continuous symmetry survives.
* If the two <eigenvalue multiplicities> are $m$ and $N-m$, then $H=S(O(m)\times O(N-m))$, with <identity component> $\operatorname{SO}(m)\times\operatorname{SO}(N-m)$.
For example a rank-one tensor $Q=ve_Ne_N^T$ preserves $O(N-1)$ embedded by $A\mapsto\operatorname{diag}(A,\det A)$. Its continuous unbroken <group> is $\operatorname{SO}(N-1)$, but it also allows reversal of the distinguished axis. Restricting to a traceless <symmetric tensor> imposes $\sum_i m_iq_i=0$ and otherwise leaves the same classification.
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