In the general RG formalism, suppose $\vec{\mu}$ represents a vector in parameter space and $\vec{\mu}^*$ is the fixed point under the transformation $R$. Then for $\vec{\mu}=\vec{\mu}^*+\delta \vec{\mu}$, we have $$R(\vec{\mu})=R(\vec{\mu}^*+\delta \vec{\mu})=\vec{\mu}^*+R'(\vec{\mu}^*)\delta \vec{\mu}$$ after which the small deviation $\delta \vec{\mu}$ is expanded as a linear combination of the eigenvectors of the matrix $R'(\vec{\mu}^*)$, denoted as $\{\vec{v}_i\}$ with $\{\lambda_i\}$ as the corresponding eigenvalues.
It seems that people normally just assume that the eigenvectors of $R'(\vec{\mu}^*)$ form a complete basis, which is obviously not true in general. Now let's assume that $\{\vec{v}_i\}$ does not form a complete basis, then we would have $$\delta \vec{\mu}=\sum_i c_i\vec{v}_i+\Delta \vec{\mu}$$ where $\Delta \vec{\mu}=\delta \vec{\mu}-\sum_i c_i\vec{v}_i$ and $\Delta \vec{\mu} \perp \vec{v}_i$ for all $i$. This simply means that $\Delta \vec{\mu}$ is the projection of $\delta \vec{\mu}$ onto the normal of the hyperplane spanned by $\{\vec{v}_i\}$. In this case, the linear transformation from the linearisation obviously keeps $\Delta \vec{\mu}$ invariant. Thus, if $\lambda_i <0$ for all $i$, we will get $\vec{\mu}^* +\Delta \vec{\mu}$ in the end, which effectively seems to be a new shifted fixed point. Indeed, we have $$R(\vec{\mu}^*+\Delta \vec{\mu})=\vec{\mu}^*+R'(\vec{\mu}^*)\Delta \vec{\mu}=\vec{\mu}^*+\Delta \vec{\mu}$$ Since $|\Delta \vec{\mu}|<<1$, it seems that the two fixed points are very close to each other and in fact they can be made arbitrarily close if the the $\delta \vec{\mu}$ is carefully chosen. This seems to be a quite strange result. Furthermore, along the way we only kept the linear terms of the expansions, whereas I suspect that higher order terms may play a role.
The overall picture seems rather blurred to me. Really appreciate if someone can help clarify.
EDIT: Thanks to @Brightsun, $\Delta \vec{\mu}$ is not necessarily invariant under the action of $R'(\vec{\mu}^*)$ (example given in the comment)