= Solution
From the derived formula,
$$
\Phi(\rho)=\frac12\left(I+(1-q)\mathbf s\cdot\boldsymbol\sigma\right),
$$
so the action on the <Bloch vector> is
$$
\boxed{\mathbf s\longmapsto(1-q)\mathbf s=\left(1-\frac{4p}{3}\right)\mathbf s.}
$$
For $0<q<1$, every direction is contracted by the same positive factor. The pure-state <Bloch sphere> becomes a concentric sphere of radius $1-q$, and the <Bloch ball> contracts towards the maximally mixed <density operator> $I/2$. The lack of a preferred direction explains the name <depolarizing channel>.
At $q=1$ all inputs become $I/2$. For the allowed Pauli-error range $1<q<4/3$, the factor is negative, so there is also an antipodal reversal; the radius is $q-1\leq1/3$. Thus it would be incorrect to describe the whole printed $p\in(0,1)$ range as positive radial contraction.
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