It well know that the isotropy group $I_q$ of $SO(3)$ (with $q\in M$ not fix point) is a Lie subgroup of $SO(3)$. Why $\dim I_q=1$?
Idea: $SO(3)$ acts transitively on the groups of orbits $\mathcal S_q$ of $SO(3)$, then $SO(3)/I_q$ must be diffeomorphic to $\mathcal S_q$. Also, $\dim \mathcal S_q=2$ and $\dim SO(3)=3$.
Can we deduce that $\dim I_q=\dim SO(3)-\dim \mathcal S_q=3-2=1$?
Since you didn't specify on which space $\operatorname{SO}(3)$ acts, your question is not really well-defined. A natural space on which $\operatorname{SO}(3)$ acts is the (unit) sphere $S^2 \subseteq \mathbb{R}^3$. The action is transitive and so $\operatorname{SO}(3) / I_q \cong S^2$ which implies that
$$ 2 = \dim S^2 = \dim \operatorname{SO}(3) - \dim I_q = 3 - \dim I_q \implies \dim I_q = 1. $$
In this case, you can of course calculate $I_q$ explicitly and see that this is indeed the case. If we choose $q$ to be the north pole $q = (1,0,0)$, then
$$ I_q = \left \{ \begin{pmatrix} 1 & 0 & 0 \\ 0 & a & b \\ 0 & b & c \end{pmatrix} \, : \, \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \operatorname{SO}(2) \right \} $$
so the isotropy group is (isomorphic to) $\operatorname{SO}(2)$ which is one-dimensional.