TITLE: Is there a global obstruction for a diffeomorphism to be an isometry?
QUESTION [12 upvotes]: Let $V$ be a finite dimensional vector space.
Let us call an automorphism $T:V\rightarrow V$ admissible if there exists an inner product $\langle , \rangle$ on $V$ making  $T$ an isometry.
We know $T$ is admissible if and only if there exists a basis for $V$ such that the representing matrix of $T$ w.r.t to this basis is orthogonal. (see this question of mine for details).
Similarly, we call a finite sequence of linear automorphisms $T_i:V_i\rightarrow V_{i+1}$ for $i=1,...n-1$ and $T_n:V_n\rightarrow V_1$ admissible if there exists inner products $\langle , \rangle_i$ on $V_i$ making all the $T_i$  isometries.
It is easy to see that a sequence is admissible if and only if the composition $T_n \circ T_{n-1} \circ ... \circ T_1:V_1\rightarrow V_1$ is admissible.
Now assume we have a diffeomorphism $\phi:M\rightarrow M$. ($M$ is a smooth manifold). A necessary condition for the existence of a Riemannian metric on $M$ making $\phi$ an isometry is a "pointwise" condition: 
For every fixed point $p\in M$ of $\phi$, $d\phi_p:T_pM\rightarrow T_pM$ should be admissible. More generally, for evey finite orbit of $\phi$ the corresponding sequence of differentials must be admissible. 
For infinite orbits we have no pointwise constraints, since we can just pick an inner product on one of the tangent spaces in the orbit, and take the induced (pulled-back or pushed-forward) inner products on consecutive spaces ad infinitum. 
My Question: Assume $\phi$ satisfies all the poinwise conditions mentioned above. Is there a Riemannian metric making $\phi$ an isometry? 
The question boils down to whether we can glue together all the different inner products into a smoothly changing inner product. (Note that we have some degrees of freedom in choosing the preserved inner product on each tangent space, since an admissible automorphism can have many different preserved products)
Update:
In the case where there are no finitie orbits (or "almost none") it is demonstrated by the answers of Matveev & Sawin below that there is a global obstruction. 
Two questions arise:
1) What happens if all the orbits are finite?
Well, it turns out that any diffeomorphism which has only finite orbits is actually of finite order, and for that case the answer is always positive, since we can sum over translates of any initial metric
2) Can we find other conditions on the dynamics of the diffeomorphism  which ensures a smooth metric exists?

REPLY [5 votes]: I have decided to write the details of the solution suggested by Sawin, for sake of completeness.
Let $M=\mathbb{S}^1\times\mathbb{R}$. (i.e $M$ is the infinite cylinder in $\mathbb{R}^3$).
$\phi:M \rightarrow M 
, \phi(p,r)=(R_{\theta}(p),2r)$. For a point $(p,x)\in \mathbb{S}^1\times\mathbb{R}$, $T_{(p,x)}(\mathbb{S}^1\times\mathbb{R})=T_p\mathbb{S}^1 \oplus T_x\mathbb{R}=T_p\mathbb{S}^1 \oplus \mathbb{R}$.  
Looking at the differential, we see that $d\phi_{(p,x)}:T_p\mathbb{S}^1 \oplus \mathbb{R}\rightarrow T_{R_{\theta}(p)}\mathbb{S}^1 \oplus \mathbb{R}, d\phi_{(p,x)}(v,r)=(R_{\theta}(v),2r)$. 
Now assume there is a $\phi$-invariant metric on $M$. 
Now define the following function: $f:\mathbb{S}^1 \rightarrow \mathbb{R}$, $f(p)=\|(0,1)\|_{(p,0)}$ (The length of a vector which points along $\mathbb{R}$, i.e, is orthogonal to the copy of $\mathbb{S}^1$ which lies at height $0$). Now note that $\phi$-invariance of the metric implies: $f(p)=\|(0,1)\|_{(p,0)}=\|d\phi_{(p,0)}(0,1)\|_{(R_{\theta}(p),0)}=\|(0,2)\|_{(R_{\theta}(p),0)}=2\|(0,1)\|_{(R_{\theta}(p),0)}=2f(R_{\theta}(p))$.
Now, since $f$ is smooth (proof in the end), it is a real-valued continuous function on a compact domain, hence obtains a maximum. Since we can rotate the point where the maximum is achieved and the value will be multiplied (or divided) by 2, this forces the maximum of $f$ to be $0$. But this is a contradiction to the positive-definiteness of the metric. 

All is left is to verify that $f$ is smooth: 
Define the vector field $X:M\rightarrow TM,$ via $X(p,x)=(0,1)\in T_{(p,x)}(\mathbb{S}^1\times\mathbb{R})=T_p\mathbb{S}^1 \oplus \mathbb{R}$. $X$ is smooth since $TM=T(\mathbb{S}^1\times\mathbb{R}) \stackrel{\text{diffeomorphic}}{\cong} T\mathbb{S}^1 \times T\mathbb{R}$ and both "components" are smooth as sections of $T\mathbb{S}^1, T\mathbb{R}$ respectively.
Now consider the function $h:M \rightarrow \mathbb{R}$, $h(p,x)=\|X(p,x)\|^2_{(p,x)}=\|(0,1)\|^2_{(p,x)}$. $h$ is smooth by the definition of Riemannian metric.
Since $\mathbb{S}^1=\mathbb{S}^1 \times \{0\}$ is a submanifold of $M$ the restriction $h|_{\mathbb{S}^1} = f^2$ is smooth. Since $f$ is positive (by definition of a metric) and $\sqrt{(\cdot)}:\mathbb{R}^{>0} \to \mathbb{R}^{>0}$ is smooth, it follows that $f$ is smooth as required.

An attempt to explain Sawin's remark about the eigenvalues:
Let $x\in M$. Assume $\phi^n(x)\rightarrow x$. We have $(d \phi^n) _x:T_xM \rightarrow T_{\phi^n(x)}M$. It is possible to choose an orthonormal local frame $E_1,...,E_n$ over a (small enough) neigbourhood $U$ of $x$. (since $\phi^n(x)\rightarrow x$ we can assume $\phi ^n(x) \in U$ for every $n$). Now assume $[d \phi^n]^{E_i(x)}_{E_i(\phi^n(x))}$ has eigenvalue $\lambda_n$, i.e there exists a vector $v_n\in T_xM$ such that 
(*) $\lambda_n [v_n]_{E_i(x)}=[d \phi^n]^{E_i(x)}_{E_i(\phi^n(x))}[v_n]_{E_i(x)}=[(d \phi^n) _x (v_n)]_{E_i(\phi^n(x))}$. 
Also, using the fact $\phi$ is an isometry implies
(**) $\|[(d \phi^n) _x (v_n)]_{E_i(\phi^n(x))}\|_{Euc} \stackrel{\text{$E_i$ orthonormal}}{=}\|(d \phi^n) _x (v_n)\|_{\phi^n(x)} \\ \stackrel{\text{$\phi$ isometry}}{=}\|v_n\|_x\stackrel{\text{$E_i$ orthonormal}}{=}\|[v_n]_{E_i(x)}\|_{Euc}$
($\| \cdot \|_{Euc}$ is the standard Euclidean norm on $\mathbb{R}^n$)
Combining (*),(**) we get: $\| \lambda_n [v_n]_{E_i(x)}\|_{Euc}=\|[v_n]_{E_i(x)}\|_{Euc}$, hence $|\lambda_n|=1$.