= Solution
Choose representatives $v_1,\ldots,v_{n+1}$ of the first $n+1$ points. The general-position condition says that these vectors form a <basis> of $\mathbb C^{n+1}$. Write a representative of the last point as $v_{n+2}=\sum_ic_iv_i$. Every $c_i$ is nonzero; otherwise the last point and the other $n$ <basis> points would lie in a <hyperplane>. Rescaling $v_i$ by $c_i$ gives representatives with $v_{n+2}=\sum_iv_i$. Normalize representatives $w_i$ of the second frame in the same way.
There is a unique invertible <linear map> $L$ with $L(v_i)=w_i$ for $i\leq n+1$. It also maps $v_{n+2}$ to $w_{n+2}$, so its projectivization has the required effect. For uniqueness, the quotient of two candidate <projective transformations> fixes the normalized frame. A representing matrix must be diagonal in the $v_i$ <basis>, with entries $a_i$. Fixing $[\sum_iv_i]$ forces all $a_i$ to be equal. Such a scalar represents the identity <projective transformation>. Thus \b[the ordered <projective frame> determines the transformation uniquely].
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