Is volume--preserving an intrinsic property? Let $M$ be a compact smooth manifold without boundary. A Riemannian metric $g$ on $M$ induces a volume measure (or Lebesgue measure) $m_g$ on $M$.
A diffeomorphism $f:M\to M$ is said to be {volume--preserving} if $f_*(m_g)=m_g$, that is, for each Borel subset $A\subset M$, $f_*(m_g)(A):=m_g(f^{-1}A)=m_g(A)$. Or equivalently the Jacobian (determinant of the tangent map $Df_x:T_xM\to T_{fx}M$) satisfies $J(f,m_g)(x)=1$ for every point $x\in M$.
If we change the Riemannian metric to $g'$ and the induced measure $m_{g'}$, the volume--preserving property with respect to $g$ is slightly distorted: there exists a uniform constant $C\ge1$ such that
$C^{-1}\le J(f^n,m_{g'})(x)\le C$, for every point $x\in M$ and every time $n\in\mathbb{Z}$.----  $(*)$
My question is: is  $(*)$ a characterization of volume--preserving? 
That is, for a given $f\in\mathrm{Diff}(M)$, if $(*)$ holds for some arbitrarily chosen Riemannian metric $g$, does there exist a Riemannian metric $g'$ such that $J(f,m_g)=1$?
Thanks!
 A: Just to add a positive result. For circle diffeomorphisms ($C^1$) one has that if the rotation number is irrational, a necessary and sufficient condition to be linearizable (by a $C^1$ map, which in particular implies you will preserve a $C^1$-length) is to have bounded derivatives (that $log |(f^n)'|$ is uniformly bounded). So, in this context you have your answer. This is a consequence of Gottshalk-Hedlund theorem, see for example the book by deMelo-VanStrien.  
A: If the volume form for $M^n$ is $\Omega$, note that the average $\Omega_k$ of
$f^{k*}(\Omega)$ for $0 \le k \le n$ is a volume form that has much tighter bounds for invariance by $f$ than $\Omega$, since all terms except the first and the
last are equal.  You can pass to the limit of the $\Omega_k$ as a measure,
and obtain an invariant measure that has the form $g * \Omega$, where $|\log(g)|$ is bounded.   
Thus, the only issue is smoothness, but this is a real issue.  Here's how to make a counterexample.  I'll describe it as a flow on $S^2$; to get a diffeomorphism, take the time one map of the flow.   Begin with the flow lines of a rotation
of the sphere about the north pole-south pole axis.  Now distort the flow lines
near a point on the equator by a non-smooth homeomorphism that squeezes the
northern hemisphere toward the point with equator, compressing say by a factor
of two near that point,  but is the identity on the
southern hemisphere.  The rotation flow conjugated by the homeomorphism is
not smooth, but multiply the image of the rotation vector field
by a function that depends only on the flow line and has $C^\infty$ contact
to 0 at the equator.  The resulting vector field $X$ is smooth.  Invariant 
measures for the flow are determined by the spacing between flow lines, so
they are equivalent to measures on $S^2$ modulo flow lines: for a smooth invariant 2-form
$\Omega'$, the identity is that the 1-form $i_X(\Omega') = 0$ would be closed. This 1-form
is 0 in the direction of $X$, that is, it would have the form $d(h)$, for some function
$h$ that is constant on flow lines.  But for any such measure, the ratio to the standard area form will have a discontinuity somewhere along the equator.
This particular example is topologically conjugate to a volume preserving flow
for a smooth Riemannian metric, but I think a little more work would give
examples that are not even topologically conjugate.
