If $F^*\alpha=\alpha$, then $F$ is the cotangent lift of some $\phi$ 
Let $Q$ be a smooth manifold, $\alpha\in\Omega^1(T^*Q)$ the tautological $1$-form and $\omega:=-d\alpha$. Suppose there is a diffeomorphism $F:T^*Q\to T^*Q$ such that $F^*\alpha=\alpha$. Prove there is a diffeomorphism $\phi:Q\to Q$ such that $F$ is the cotangent lift of $\phi$. Suggested plan:
1) Prove that the Euler field $X$ (defined by $i_X\omega =-\alpha$) is invariant under $F$, i.e., $dF(X)=X\circ F$.
2) If $\phi_t^X$ is the flow of $X$, prove that $\phi_t^X\circ F=F\circ\phi_t^X$ and that $\phi_t^X(x,\xi)=(x,e^t\xi)$ locally.
3) Verify that $F(\lambda m)=\lambda F(m)$ for all $m\in T_q^*Q$, $\lambda\in\mathbb{R}$. Conclude there exists $\phi$ such that $\phi\circ\pi=\pi\circ F$ and that $F$ is the cotangent lift of $\phi$.

I've already gone through $1)$ and $2)$ but I'm stuck with $3)$. 
Here where I'm at: for $\lambda>0$, we have:
$$F(\lambda m)=F(x,\lambda\xi)=F(x,e^{ln(\lambda)}\xi)=F\circ\varphi_{ln(\lambda)}^X(x,\xi)=\varphi_{ln(\lambda)}^X\circ F(x,\xi)=\lambda F(m)$$
By continuity, that also covers the case $\lambda=0$, but I don't know what to do when $\lambda<0$. I tried doing $F(\lambda p)=F(-\lambda(-p))=-\lambda F(-p)$, but I can't prove $F(-p)=-F(p)$.
I know how to prove that if such a $\phi$ exists it must be a diffeomorphism. 
My problem is to define $\phi$ and to check $F$ is its cotangent lift.
If I could prove $F$ is linear on the fibers, then I could just define $\phi:=\pi\circ F\circ \eta$, where $\eta:Q\to T^*Q$ is any $1$-form in $Q$, but I wasn't able to prove the aditive property (i.e., $F(v+w)=F(v)+F(w))$. Besides, the exercise appararently suggests we only need the scaling property, which is also confusing me.
For $F$ being the cotangent lift, I'm also stuck.
 A: The facts (1) and (2) suffice to show that a map $\phi : Q \to Q$ satisfying $\pi \circ F = \phi \circ \pi$ exists, as $F$ sends whole fibers to whole fibers. (Remark: using the zero section and the assumption that $F$ is a diffeomorphism, it follows that $\phi$ is a diffeomorphism.)
Let $q \in Q$, $v \in T^*_q Q$ a nonzero covector and consider the 1-dimensionnal vector subspace $L =  \mathbb{R}\langle v \rangle \subset T^*_q Q$. By assumption $F$ is a diffeomorphism, so $F(L) \subset F(T^*_q Q) = T^*_{\phi(q)}Q$ is a smooth line; in particular $F(L)$ has a tangent line at $0 \in T^*_{\phi(q)}Q$. As you observed the fact (2) shows that $F(L)$ is $\mathbb{R}_+$-homogeneous, so the previous sentence implies that $F(L)$ is a genuine straight line. A priori this only implies that $F(q,-v) = - tF(q,v)$ for some $t = t(v/|v|)$ possibly different from $1$, but consideration of the linearity of the restricted map $DF_{(q,0)} : T_0(T^*_q Q) \to T_0(T^*_{\phi(q)}Q)$ shows that $t=1$. We thus get that $F(q,-v) = -F(q,v)$ for all $(q,v) \in T^*Q$, and thus the first part of (3).
We know that the vector bundle $T^*Q$ is canonically isomorphic to the restriction to the zero section $0_Q$ of the vertical bundle $V := \mathrm{Ker}(d\pi)$. Using what we just established, it turns out that this isomorphism intertwines $F$ and the restriction of $DF$ to this bundle $\left. V \right|_{0_Q}$; in particular $F$ is linear on fibers. But this restriction of $DF$ is the cotangent lift of the restriction of $F$ to $0_Q$, that is the cotangent lift to $\phi$. Hence $F$ and the cotangent lift of $\phi$ are two bundle morphisms which are intertwine with the same map on $\left. V \right|_{0_Q}$, hence they are equal.
There are a lot of 'tautological' identifications involved in the previous paragraph. Alternatively, let $\Phi$ denote the cotangent lift of $\phi$ and set $P = F \circ \Phi^{-1}$; we therefore wish to prove that $P = id_{T^*Q}$. Observe that $P$ is a diffeomorphism, that $P^*\alpha = \alpha$ and that $\pi \circ P = id_Q \circ \pi$. Fix $(q,v) \in T^*Q$ and chosse a section $\eta : Q \to T^*Q$ such that $\eta(q) = (q,v)$. Define $\tilde{\eta} = P \circ \eta$. Now observe that (this is the 'tautological part' of the argument)
$$ \tilde{\eta} = \tilde{\eta}^*\alpha = \eta^* P^* \alpha = \eta^* \alpha = \eta \, , $$
which implies (since $(q,v)$ was arbitrary) that $P = id_{T^*Q}$.
