We seek Clifford numbers c such that φ(c) = cvc−1 leaves all elements v of V fixed, i.e. v = cvc−1 (↔ cv = vc). The only such numbers I know now are the non-zero reals within Cl(n). We restrict ourselves to values v such that (ip v v) = 1. It suffices to require that b = cbα(c)−1 for each base element b of V.

An alternate definition of the Clifford group uses the expression cvα(c)−1 in place of cvc−1. This defines the same set of Clifford numbers but the orthogonal matrices they lead to have the opposite sign when c is odd.

To find those Clifford numbers c that commute with each member of V we first seek all c such that cγ0 = γ0c.
Split c thus c = f + γ0g where neither f nor g mention γ0.
0 = γ0c ↔ (f + γ0g)γ0 = γ0(f + γ0g) ↔ fγ0 + γ00 = γ0f + γ0γ0g ↔
γ0f = fγ0 & γ00 = γ0γ0g ↔ f and g are both pure even. (Note that γ0 has an inverse: −γ0.)
The code corroborates this.

Said another way, a Clifford number commutes with γ0 iff each of its terms mentions an even number of other base vectors. When we require that a Clifford number commute with each of the base elements we are clearly left with only the reals among the Clifford numbers.