Projective equivalence of two sets of $n+3$ points in $\Bbb P^n$ and on the rational normal curves through each of them We know that through any $n+3$ points in general position in $\mathbb P^n$ ($n$-dimensional projective space) there is a unique rational normal curve. Let $p_1,p_2,\ldots,p_{n+3}$ be such points and let $q_i$ be their images on the unique rational normal curve through all the $p_i$. Now we are asked to show that $(p_1,p_2,\ldots,p_{n+3})$ is projectively equivalent as an ordered set to another such collection $(p_1',\ldots,p_{n+3}')$ iff the corresponding ordered subsets $(q_1,\ldots,q_{n+3})$ and $(q_1',\ldots,q_{n+3}')$ in $\mathbb P^1$ are projectively equivalent, that is, iff the cross ratios $(q_1,q_2,q_3,q_i)=(q_1',q_2',q_3',q_{i}')$ for each $i = 4,\ldots,n+3 $.
This is a problem from page $12$ of Harris's first course book.

We know that any n+2 points in general position in $\mathbb{P}^n$ are projectively equivalent, so without loss of generality we can take $\{q_1, q_2, q_3\}$ to be (projectively equivalent) to $\{0,1,\infty\}$ via a unique $\Phi \in PGL_2(k)$, and the cross ratio $\lambda(q_1,q_2,q_3,q_4)$ determines the image $\Phi(q_4)$. Furthermore, we already know that the ordered subsets $\{p_1, \dots, p_{n+2}\}$ and $\{p_1', \dots, p_{n+2}'\}$ are projectively equivalent since they are in general position in $\mathbb{P}^n$. The question now simplifies to what happens when we add in $p_{n+3}$ and $p_{n+3}'$, respectively.
We can consider $\lambda(q_1,q_2,q_3,q_i)$ for $4\leq i \leq n+3$ to be the image  $\Phi(q_i)$ rewritten in the new basis where $\{q_1,q_2,q_3\}$ are taken to be $\{0,1,\infty\}$. If the corresponding cross ratios match, then $\{q_1, \dots, q_{n+3}\}$ and $\{q_1', \dots, q_{n+3}'\}$ are projectively equivalent in $\mathbb{P}^1$. To find the projective transformation of $\mathbb{P}^n = \mathbb{P}(k^{n+1})$, let $\Gamma = \{p_1, \dots, p_{n+3}\}$ so that $\mathcal{B}=\{p_1, \dots, p_{n+2}\}$ is a projective basis for $\mathbb{P}^{n}$. Choose non-zero representatives $v_i \in p_i$. Then $v_1, \dots, v_{n+1}$ span $k^{n+1}$ so that we can write, for $v_{n+2}$ a representative of $p_{n+2}$ $$ v_{n+2}= \sum_{i=1}^{n+1}\lambda_i v_i $$
and re-scale the $v_i$ to set $\tilde{v_i}:= \frac{1}{\lambda_i}v_i$ for $1\leq i \leq n+1$. Now $$ [\phi_\mathcal{B}] = \left[\begin{matrix} \uparrow & & \uparrow \\ \tilde{v_1} & \dots & \tilde{v_{n+1}} \\ \downarrow & & \downarrow \end{matrix}\right] \in PGL_{n+1}(k) $$
satisfies $\phi(e_i) = \tilde{v}_i$ for $1\leq i \leq n+1$ and $e_i$ standard basis vectors for $k^{n+1}$. Now, $\phi(\sum_{i=1}^{n+1} e_i) = v_{n+2}$ by construction, from the re-scaling. So, it remains to find $[\phi_\mathcal{B}^{-1}(p_{n+3}) ] \in \mathbb{P}^n$. If we repeat the same process with $\Gamma' = \{p_1', \dots, p_{n+3}'\}$ and set $\mathcal{B}' = \{p_1', \dots, p_{n+2}'\}$ to get $[\phi_{\mathcal{B}'}^{-1}] \in PGL_{n+1}(k)$ satisfying $\mathcal{B}' \mapsto \{e_i : 1\leq i \leq n+1\}$ then we simply ask $$[\phi_\mathcal{B}^{-1}(p_{n+3})] \stackrel{?}{=} [\phi_{\mathcal{B}'}^{-1}(p'_{n+3})] $$
And this is equivalent to the corresponding cross-ratios in $\mathbb{P}^1$ being equal, since $\nu_n\circ \Phi$ is the unique rational normal curve through the $d+3$ points $\nu_d\circ \Phi(q_i) = \nu_d\circ \Phi(q_i')$ for $1\leq i \leq d+3$ which is now projetively equivalent to both the unique rational normal curve through $\Gamma$, and to the unique rational normal curve through $\Gamma'$.
