If a metric has a particular symmetry, will a perturbation of that metric enjoy the same symmetry? By perturbation of a metric, I mean
$g_{\mu \nu}^{perturbed} = g_{\mu \nu} + h_{\mu \nu},$
where $h_{\mu \nu}$ is the perturbation with components which are small in some suitable norm.
I am assuming the answer is no: for example, the Schwarzchild metric is spherically symmetric and remains invariant under rotations (the isometry group of Schwarzchild spacetime contains a subgroup which is isomorphic to $SO(3)$ and the orbits of that subgroup are $2$-spheres), but one could directly construct some perturbation of the Schwarzchild metric such that the metric as a whole is no longer invariant under rotations.
Certainly not, unless the symmetry (which here I interpret as "isometry") is the identity, which preserves every object.
If an isometry $\phi : M \to M$ of a Riemannian metric $(M, g)$ is not the identity, there is some $p \in M$ such that $\phi(p) \neq p$. Since $\phi$ is an isometry, $g_p = (\phi^* g)_p = \phi^* g_{\phi(p)}$, so if we perturb $g$ to another metric $g'$ in a way such that $g'_{\phi(p)} \neq g_{\phi(p)}$ but $g'_p = g_p$, then by construction $\phi$ will not be a symmetry of $g'$.
Given a smooth, $1$-parameter family $\phi_t$ of isometries $(M, g) \to (M, g)$ and a bilinear form $h \in S^2 T^* M$, the metric $g' := g + h$ satisfies $$ \left.\frac{d}{dt} \phi^* g'\right\vert_0 = \mathcal L_X g' = \mathcal L_X g + \mathcal L_X h = \mathcal L_X h , $$ so $\mathcal L_X h$ is a measure of the failure for $\phi_t$ to preserve $g'$. In particular, if $X \neq 0$ (and at least when $\dim M > 0$), there is always some $h$ such that $\phi_t$ is not a symmetry of $g'$ for $t$ sufficiently closed to $0$.