Is there a slick way to show that finite projective planes of $7$ points are unique up to isomorphism?

It seems you can use the enormous symmetry of the Fano plane for this. You can pick a line $ABC$ and a point off the line $D$ and pick $f(A)$ and $f(B)$ arbitrarily, $f(C)$ is the third point on the line $AB$, and pick $f(D)$ arbitrarily different from the first three. Then if the lines through $D$ are $ADE$, $BDF$, and $CDG$ you can define $f(E)$, $f(F)$ and $f(G)$ in the obvious way. Now you just need to prove that $f(CEF)$, $f(AGF)$, and $f(BEG)$ are lines. But there has to be a meet between the lines $f(CE)$ and $f(AG)$ and it can't be $f(B)$ or $f(D)$, so must be $f(F)$.

Added: support for this comes from that fact that any isomorphism should be able to be composed with all the automorphisms of the destination plane to make another. This isomorphism has $7*6*4$ possibilities as the first, second, and fourth points can be chosen at random and $7*6*4=168$ is the number of automorphisms of the Fano plane.

Let $P$ be a projective plane on $7$ points. Fix an ordered $4$-tuple of distinct points $(p_1,p_2,p_3,p_4)$ in $P$ such that no line is incident with more than two of them. If $\pi=\{\{u,v\},\{w,t\}\}$ is a partition of $\{1,2,3,4\}$ in two parts of size $2$, let $q_\pi$ be the point of intersection of the line joining $p_u$ and $p_v$ with the line joining $p_w$ and $p_t$.

This provides a labeling $p_1$, $p_2$, $p_3$, $p_4$, $q_{\{\{1,2\},\{3,4\}\}}$, $q_{\{\{1,3\},\{2,4\}\}}$, $q_{\{\{1,4\},\{2,3\}\}}$ of the $7$ points.

Now show from the labeling alone you can reconstruct the lines.

This proves uniqueness.

NB: Notice that there are the $4$-element set $\{p_1,p_2,p_3,p_4\}$ is determined by its complement, which is a line. It follows that there are $7$ possible sets, one for each line, and each such set can be turned into an ordered $4$-tuple in $4!$ ways. The above reasoning shows then that the automorphism group of $P$ has order $4!\cdot 7=168$.