On the proof about the dimension of the conformal group of a manifold I have been reading the book "Transformation Groups in Differential Geometry" by S. Kobayashi. More concretely, I am trying to understand the proof of the Theorem 6.1 of Chapter IV.
Theorem 6.1 (S. Kobayashi, 1954): Let $M$ be a manifold of dimension $n \geq 3$, and $P$ a conformal structure on $M$. Then, the conformal transformation group  Conf$(M)$ of $M$ is a Lie group with dimension $\leq \frac{1}{2}(n+1)(n+2)$.
I think I understood the essence of the proof, which is the following: a conformal structure $P$ on $M$ is determinated by a (unique) Cartan connection on $P$ that "comes from" the restriction of the canonical form of second frame bundle $P^2(M)$ to $P$ (that connection is constructed in the section 5 of the same chapter and extended to a Cartan connection on $P$ thanks to Theorem 4.2).
The uniqueness of that Cartan connection holds since $n\geq 3$. 
Now, by using the Theorem 3.1 (of the same book), that states that the group $U(P,\omega)$ of automorphisms of $P$ that preserve the Cartan connection $\omega$, is a Lie group and has dimension $\leq \dim P = \frac{1}{2}(n+1)(n+2)$, we obtain the result.
Unfortunately, there are two details that I cannot understand. That is why I am posting this question.
(1) In the page 143, it is written "Assume that $n\geq 3$, so that the normal Cartan connection is unique. Then, for each automorphism $f$ of $P$, $f_*$ (restricted to $P$), preserves the normal connection."
I suppose that this means that for any conformal transformation $f$ on $M$, the induced map $f_*: P^2(M) \rightarrow P^2(M)$ that sends $P$ into $P$ preserves the Cartan connection $\omega$ on $P$. Thus, we have that any conformal transformation is an automorphism of $U(P,\omega)$ and then Conf$(M) \subset U(P, \omega)$. However this affirmation is not clear for me. In other words, how do we know that if the Cartan connection is unique, then any automorphism of $P$ preserves that connection?
(2) The other detail that intrigues me is the following: Do we have the equality Conf$(M) = U(P, \omega)$ in general or just Conf$(M) \subset U(P, \omega)$?
By reading the book it seems that the equality holds; that should justify the fact that Conf$(M)$ is a Lie group by using Theorem 3.1, but I still can't see how to prove this fact.
Any comment is highly appreciated.
 A: I don't have the book at hand, and I am not sure wheher the interpretation of the construction you give is really correct. (I would think that only obtains a Cartan connection on the subbundle $P$ and not on the full second order frame bundle, but actually this is not really relevant for the answer.) Also, I assume that the meaning of "automorphism $f$ of $P$" includes that $f$ preserves the soldering form, so it is indeed induced by a conformal isometry of $M$.
Then the answer to (1) basically is that given the map $\tilde f:P\to P$ induced by $f$ and $\omega$, the pullback $\tilde f^*\omega$ also is a Cartan connection on $P$, which has all the properties of $\omega$ that are required for the uniquness statement. (This partially follows from the construction of $\tilde f$ and partly from the fact that the curvature of $\tilde f^*\omega$ is the pullback of the cuvature of $\omega$.) Then indeed uniqueness implies that $f^*\omega=\omega$.
For (2) you actually do have equality. Suppose that $F:P\to P$ is a principal bundle map such that $F^*\omega=\omega$. Then one deduces that $F$ is the restriction to $P$ of the prolongation of its base map $f:M\to M$. Since this prolongation preserves $P$, it follows that $f$ is a conformal ismoetry.
