I am just adding a few details to Vladimir's answer:

**Lemma:** Assume there exists a point $x$ such that the orbit w.r.t. the iterations of the $\phi$ (i.e., the set $\{x,ϕ(x),ϕ(ϕ(x)),...\}$ is dense in $M$. Then any $g \in Met(\phi)$ is completely determined by its restriction $g_x$ to $T_xM$.

**Corollary:**  
The cone of metrics preserved by $ϕ$ is finite-dimensional. (In fact its dimension is bounded above by $\frac{n(n+1)}{2}$ which is the dimension of the manifold of all inner products on an $n$-dimensional vector space).

**Proof of lemma:**  
Let $g \in Met(\phi)$. Take $y \in M$. By the density assumption, it follows that the there exist a sequence $n_k \in \mathbb{N}$, such that $\phi^{n_k}(x)$ converges to $y$. Take a coordinate neighbourhood around $y$. Then continuity of the metric implies: $g_{ij}\big(\phi^{n_k}(x)\big) \rightarrow g_{ij}(y)$. However, $\phi^{n_k} \in \text{Iso(M,g)} \Rightarrow g_x$ determines $g_{\phi^{n_k}(x)}$ so we are done.

**Question:** Can we choose the metric $g_x$ arbitrarily?