= Solution
For a <projective frame>, choose representative vectors $v_1,\ldots,v_{n+1}$ forming a <basis>. Express the final point by $\sum c_iv_i$. Every $c_i$ is nonzero, since otherwise some $n+1$ of the given representatives would be dependent. Rescale the first representatives to $c_iv_i$, so the last point is represented by their sum. Do the same for the target frame, with <basis> $w_i$. The <linear map> taking each rescaled source <basis> vector to the corresponding target vector induces the required <projective transformation>. For uniqueness, a transformation fixing the first $n+1$ coordinate points is represented by a diagonal matrix. Fixing their sum forces every diagonal entry to agree. It is therefore scalar, giving \b[a unique <projective transformation>].
If three intersection points of the conics lay on a line, Bézout's theorem would force that line to be a component of either nonsingular <conic>, impossible. Thus the four points form a <projective frame> and can be moved to the three coordinate points and $[1:1:1]$. Vanishing at the coordinate points removes the square terms from each <conic> equation; vanishing at the last point gives $a+b+c=0$.
Substitution gives, by direct expansion,
$$
ayz+bzx+cxy=au^2+bv^2+cw^2+(a+b+c)(uv+uw+vw)=au^2+bv^2+cw^2.
$$
The substitution matrix
$$
S=\begin{pmatrix}0&1&1\\1&0&1\\1&1&0\end{pmatrix}
$$
has determinant two, hence is invertible over $\mathbb C$. If the initial frame transformation is represented in the direction old coordinates equal $Q$ times new coordinates, take $P=QS$. Both quadratic forms then have no mixed terms, proving \b[$P^TAP$ and $P^TBP$ are simultaneously diagonal]. The same argument applies when an individual <conic> is reducible but the intersection consists of exactly four distinct points: a line containing three would be a common component, contradicting finiteness of the intersection.
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