I know that this question is trivial for some, but everywhere I look, the authors already assume that the reader knows what it is about. The definition of Clifford group is a set of invertible elements $Cl^{\times}(V, Q)$ in Clifford algebra $Cl(V, Q)$ which "commutes" with the elements of vector space $V$,
$$P(V, Q) := \{x \in Cl^{\times}(V, Q)| \ x v x^{-1} \in V, \forall v \in V\}.$$ I ain't sure about what this means.
My attempt:
In definition of Clifford algebra, some author defined it to be a pair $(Cl(V, Q), f)$ with $f: V \rightarrow Cl(V, Q)$ a linear map. I think that
$$x v x^{-1} ,$$ means
$$x \cdot f(v) \cdot x^{-1},$$ where $\cdot$ is the Clifford multiplication (required in the algebra definition).
Is my approaches correct?
Appreciate.
Your approach is correct. It should be noted that writing $xvx^{-1} = v$ instead of $x \cdot f(v) \cdot x^{-1} = f(v)$ is fairly standard abuse of notation, and comes from the idea that we should view $V$ as a subspace of $Cl(V)$.