Let $\phi$ be a homogeneous symplectomorphism of tangent bundle $\dot{T}^\*M=T^\*M-0\_M$ and let $\alpha\_M$ be the canonical Liouville 1-form of $\dot{T}^\*M$. Then is it true that $\phi^\*\alpha\_M=\alpha\_M$?



Any homogeneous symplectomorphism of cotangent bundle $\dot{T}^*M=T^*M-0_M$ preserves the canonical Liouville form?
Yes this is true. You can prove it in Darboux coordinates $(x,\xi)$.
Let $\phi=(\phi\_1,\phi\_2)$ be the symplectomorphism. Since it is homogeneous, $\phi(x,t\xi) = (\phi\_1(x,\xi), t\phi\_2(x,\xi))$ and differentiating wrt $t$ you get


$$ \partial\_\xi\phi\_1 \cdot \xi = 0$$
and
$$ \partial\_\xi\phi\_2 \cdot \xi = \phi\_2$$
(these are Euler identities).


On the other hand, $\phi$ is a symplectomorphism, so $d\phi\_2 \wedge d\phi\_1 = d\xi \wedge d x$. Applying these 2-forms to a pair of vectors $(U,V)$ of the form
$$U=(0,\xi) \text{ and } V=(v,0)$$
you get $$d\xi \wedge d x (U,V) = \xi\cdot v $$ 
while 
$$d\phi\_2 \wedge d\phi\_1 (U,V) = (\partial\_\xi \phi\_2 \cdot \xi)(\partial\_x \phi\_1 \cdot v) - (\partial\_\xi \phi\_1 \cdot \xi)(\partial\_x\phi\_2\cdot v).$$
The equality between these 2-forms, combined with the Euler identities, gives
$$\phi\_2 \partial\_x\phi\_1\cdot v = \xi\cdot v,\quad \forall v,$$
which is the same as
$$\phi^\* (\xi d x) = \xi d x.$$
So the Liouville 1-form $\xi d x$ is preserved.


**Alternative, coordinate free proof:** you want to use Bertram Arnold's claim, which can be proved as follows: (since you were asking whether $\alpha$ was homogeneous). 


1. We denote the dilation action on the cotangent fibers by $h\_t(\beta) =e^t \beta$, where $\beta$ is a 1-form on $M$. Differentiating with respect to $t$, you see that $h\_t$ is the flow of the Euler vector field, which is tangent to the vertical fibers: $E(\beta) = (0,\beta)$. Hence $\iota\_E \alpha = 0$.
2. Now there is a nice characterization of the Liouville 1-form by the tautological formula $\beta^\* \alpha = \beta$, for any 1-form $\beta$ on $M$ viewed as a map $M\to T^\*M$. Hence $\beta^\*(h\_t^\*\alpha) = (h\_t\circ\beta)^\*\alpha = h\_t\circ\beta = e^t\beta$, which implies that $h\_t^\*\alpha = e^t \alpha$.
3. You differentiate this last identity at $t=0$ to obtain
$$\mathcal{L}\_E\alpha = \alpha.$$
The Cartan formula together with (1.) yields $\iota\_E d\alpha = \alpha$.
4. Let $\phi$ be a homogeneous symplectomorphism: $\phi\circ h\_t = h\_t\circ \phi$, and $\phi^\*(d\alpha) = d\alpha$. From the first equality you obtain that the operators $\iota\_E$ and $\phi^\*$ commute. Hence


$$ \phi^\*\alpha = \phi^\*(\iota\_E d\alpha) = \iota\_E \phi^\* (d\alpha) = \iota\_E d\alpha = \alpha. $$


