$\DeclareMathOperator{\vol}{vol}\newcommand{\Bar}[1]{\overline{#1}}$tl; dr: Yes, $M$, and $N$ are isometric, assuming only that each is connected, complete, and of finite volume.
Lemma 1: If $(M, g)$ is a complete Riemannian manifold, $(N, h)$ is a connected Riemannian manifold, and $\dim M = \dim N$, then an isometric immersion $i:(M, g) \to (N, h)$ is a surjective covering map.
Proof: The image $i(M)$ is open (because $i$ is an isometric immersion, hence a local diffeomorphism) and closed (because $(M, g)$ is complete) and non-empty, hence equal to $N$ (since $N$ is connected).
Let $(\Bar{M}, \Bar{g}) = (M, g)/i$ denote the Riemannian quotient. That is, define an equivalence relation on $M$ by $p \sim p'$ if and only if $i(p) = i(p')$. Since $i$ is an isometric immersion and $\dim M = \dim N$, the quotient acquires the structure of a smooth Riemannian manifold isometric to $(N, h)$. Let $\pi:M \to \Bar{M}$ denote the quotient map.
Let $q$ be an arbitrary point of $\Bar{M}$, and $V_{r} = V_{r}(q) \subset (\Bar{M}, \Bar{g})$ the geodesic ball of radius $r$ about $q$. Fix a point $p \in \pi^{-1}(q)$ arbitrarily, and choose $r > 0$ small enough that $U_{r}(p) \subset (M, g)$, the geodesic ball of radius $r$ about $p$, is mapped isometrically to $V_{r}$ by $\pi$.
To complete the proof, it suffices to show that $\pi^{-1}(V_{r})$ is a disjoint union of geodesic balls, each mapped isometrically to $V_{r}$ by $\pi$. With the notation of the preceding paragraph, $U_{r}(p) \subset \pi^{-1}(V_{r})$. Conversely, if $x$ is a point of $\pi^{-1}(V_{r})$, so that $\Bar{x} = \pi(x) \in V_{r}$, there is a minimal geodesic $\Bar{\gamma}$ joining $\Bar{x}$ to $q$. Since $\pi$ is a local isometry, the geodesic $\gamma$ that starts at $x$ and satisfies $\pi_{*}\gamma'(0) = \Bar{\gamma}'(0)$ is a lift: $\Bar{\gamma} = \pi \circ \gamma$. Consequently, $\gamma$ joins $x$ to some point $p$ in $\pi^{-1}(q)$. Since $d(x, p) = d(\Bar{x}, q) < r$, we have $x \in U_{r}(p)$.
Lemma 2: If $(M, g)$ and $(N, h)$ are Riemannian manifolds with $(M, g)$ connected, complete, and of finite volume, and if there exist isometric immersions $i:M \to N$ and $j:N \to M$, then $j \circ i:M \to M$ is an isometry
Proof: Suppose $i:M \to N$ and $j:N \to M$ are isometric immersions. (Particularly, $\dim M = \dim N$.) The composition $j \circ i:M \to M$ is an isometric immersion, hence by Lemma 1 a covering map, say with $d$ sheets, so that $\vol(M) = d\vol(M)$. Since $\vol(M)$ is finite, $d = 1$. That is, $j \circ i$ is a diffeomorphism as well as a local isometry, hence an isometry.
Corollary: If $(M, g)$ and $(N, h)$ are complete, connected, finite-volume Riemannian manifolds, and if there exist isometric immersions $i:(M, g) \to (N, h)$ and $j:(N, h) \to (M, g)$, then $i$ and $j$ are isometries.
Proof: By Lemma 2, $j \circ i$ is bijective, so $j$ is surjective and $i$ is injective. Reversing roles, $i \circ j$ is bijective, so $i$ is surjective and $j$ is injective. That is, each of $i$ and $j$ is a bijective isometric immersion, hence an isometry.
Best Answer
The basic idea is the following.
For each point $p\in N$, the fiber $f^{-1}(p)$ is a set of $n$ points in $M$. Each such point has a neighbourhood such that the restriction of $f$ to it is an isometry. Now take the intersection of the images of these sets, which will be some open neighbourhood $p \in U \subset N$. By its definition, the restriction of $f$ to each connected component of $f^{-1}(U)$ will be an isometry. I claim you can prove the following (let $\nu$ denote the volume form on $N$ corresponding to the metric):
Lemma
$$Vol(f^{-1}(U)) = \int_{f^{-1}(U)}f^*(\nu) = n\cdot \int_{U}\nu = n\cdot Vol(U)$$
In fact, we have the following slightly stronger statement:
Lemma For any function $\psi$ and a volume form $\nu$,
$$\int_{f^{-1}(U)}f^*(\psi\nu) = n\cdot \int_{U}\psi\nu $$
This lemma is a local version of the statement you want. We used the local isometry property here. We would now like to make it global. How do you typically achieve such a thing in differential geometry? One common strategy is to use a partition of unity.
Take a cover of $N$ by open sets $U_\alpha$, such that for each $U_\alpha$, the restriction of $f$ to each connected component of $f^{-1}(U_\alpha)$ is an isometry. (I claim you can do this, but you should check carefully that it is true).
Take a partition of unity $\{ \psi\}$ on $N$ that is subordinate to your chosen open cover (I omit indices). The volume form for our Riemannian metric, $\nu$, is equal to
$$ \nu = \sum \psi \cdot \nu.$$
(where the sum is over functions in our partition). Now
$$ Vol(M) = \int_M f^*\nu = \int_M f^*\left(\sum \psi \cdot \nu\right) = \sum\int_M f^* \left(\psi \cdot \nu\right) = \sum\int_{f^{-1}(U_\alpha)} f^* (\psi \cdot \nu)$$
(if your volume is infinite there might be some convergence issues around commuting the sum and integral, but this case can probably be handled separately). By our lemma,
$$ \int_{f^{-1}(U_\alpha)} f^* \psi \cdot \nu = n \cdot \int_{U_\alpha} \psi \cdot \nu $$
Putting this together with the fact that $\psi$ is a partition of unity, we continue from above:
$$\sum\int_{f^{-1}(U_\alpha)} f^* (\psi \cdot \nu) = n\sum \int_{U_\alpha} \psi \cdot \nu = n \int_{M} \sum\psi \cdot \nu =n \int_{M} \nu = n Vol(N)$$