Given a conic $C$ is the map For $A\in C,$ $A\to AX\cap C$ Where we take the point which isn't $A$.(except when $AX$ is a tangent) a projective transformation from the conic to itself? where $X$ is a fixed point not on $C$
I'm pretty sure the statement is correct but I've been unable to find a source for it.
I have a proof sketch but I'm unsure whether it is actually true:
Project $C$ to be a circle and $X$ it's center, then the transform is just a rotation which is projective.
Is this argument true?
Yes, your proof is fine.
For an alternative proof note that the mapping $T$ you give extends to the entire plane. For any point $P$, take any two lines $\ell, m$ passing through $P$ that are secants to $C$. You can find the intersections of the secants with $C$ and use the mapping to find lines $\ell',m'$ whose intersection gives $F'=T(F)$.
$T$ is a perspective mapping whose center is $S$ and whose axis is the polar of S. So $C$ is in perspective with itself.
This is covered in lots of old (19th and early 20th century) textbooks, e.g. Hatton's Projective Geometry, pg 173 (o). That reference refers back to Article 79 in the same text.