= Solution
For the <Möbius transformation> $z'=(az+b)/(cz+d)$ with $ad-bc=1$,
$$
z_j'-z_k'=\frac{z_j-z_k}{(cz_j+d)(cz_k+d)},
\qquad
d^2z_i'=\frac{d^2z_i}{|cz_i+d|^4}.
$$
The power of $|cz_i+d|$ contributed by every pair containing $i$ is
$$
-\alpha'\sum_{j\ne i}p_i\cdot p_j
=\alpha'p_i^2=4,
$$
where <momentum conservation> and the mass-shell condition $p_i^2=4/\alpha'$ were used. The Koba-Nielsen factor therefore contributes $|cz_i+d|^4$ at each insertion, exactly cancelling the transformed measure. Hence the remaining integral is $SL(2,\mathbb C)$ invariant, and division by its volume removes the residual conformal-gauge redundancy.
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