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?
Some details concerning the example on the torus:
Let $N \in \mathbb{N}$. By Dirichlet's approximation theorem, there exist integer numbers $1 \le q \le N^2,p$ such that: $|q\alpha_1-p|\le \frac{1}{N^2+1}$. Now, applying the theorem again (now for $q\alpha_2$) we can obtain $1 \le q' \le N,p'$ such that: $|q'(q\alpha_2)-p'|\le \frac{1}{N+1}$. So,$|q'q\alpha_1-q'p|\le N \cdot \frac{1}{N^2+1} \le \frac{1}{N}$. Denoting $qq'=m,q'p=n_1,p'=n_2$ we get: $|m\alpha_1-n_1|\le \frac{1}{N},|m\alpha_2-n_2|\le \frac{1}{N}$, so putting $\epsilon_1 = m\alpha_1-n_1,\epsilon_2 = m\alpha_2-n_2$ we get $g_{ij}(x,y)=g_{ij}(x+\epsilon_1,y+\epsilon_2)$ for any $(x,y)$. Hence $g_{ij}$ is constant. (Since it's continuous and has arbitrarily small peroids).
(Where we used the irrationality of $\alpha_i$ in the fact that the obtained periods $\epsilon_i \neq 0)$