= Solution
For a fixed point $g_f$, put $g=g_f+\delta g$. Linearizing the <renormalization-group flow> gives
$$
\frac{d\delta g}{d\ell}=\mathcal B'(g_f)\delta g+O(\delta g^2),\qquad
\mathcal B'(g_f)=-\epsilon+2(n-2)A_Dg_f.
$$
Because $g$ is proportional to temperature at fixed microscopic stiffness, this derivative is the thermal eigenvalue. The <thermal eigenvalues of the spherical nonlinear sigma model> are
$$
\boxed{y_T(0)=-\epsilon,\qquad y_T(g_*)=\epsilon\quad(n>2,\ \epsilon>0).}
$$
The first value describes the stable zero-temperature ordered fixed point. At the nonzero critical point the thermal perturbation grows as $e^{\epsilon\ell}$, so the stopping scale gives $\xi\propto|g-g_*|^{-1/\epsilon}$ and $\nu=1/\epsilon$ to leading order in the expansion. Higher-loop terms change this to $y_T=\epsilon+O(\epsilon^2)$.
At $\epsilon=0$ the linear eigenvalue is zero, so its sign alone does not decide the flow. For $n>2$ the first nonlinear term is positive. Integrating it gives
$$
\frac1{g(\ell)}=\frac1{g(0)}-\frac{n-2}{2\pi}\ell.
$$
The flow reaches order-one coupling at a length logarithm of order $2\pi/[(n-2)g(0)]$, explaining the <exponentially large correlation length in a two-dimensional spherical sigma model> rather than a finite-temperature power-law critical point in this smooth-field regime. For $n=2$, $\epsilon=0$, the linear eigenvalue is zero everywhere on the perturbative line and vortex effects require an additional coupling. For $n=2$, $\epsilon>0$, the sole zero-coupling fixed point has $y_T=-\epsilon$.
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