2) You are wrong in that admitting a certain structure is not equivalent to having (fixed choice of) certain structure. In other words, for any smooth manifold $M$ there is a Riemannian manifold $(M,g)$, such that if you forget the metric $g$ on $M'$, you will get just $M$. The catch is that the pair $(M,g)$ is not unique.
As a silly example, take the sphere $S^2$. You can embed it into $\mathbb{R}^3$ as the unit sphere to get one (induced) metric on it, or as a cube (smoothed), and get another induced metric. But as smooth manifolds (forgetting the metric), those are the same.
As a more interesting example, take the flat torus $\mathbb{R}^2/\mathbb{Z}^2$ with the induced metric from $\mathbb{R}^2$, and the torus which is the surface of a doughnut, embedded into $\mathbb{R}^3$. As smooth manifolds they are the same, but as Riemannian manifolds they differ; for example one can be (indeed is) embedded into $\mathbb{R}^3$, and the other one cannot.
An attempt of proof, please review carefully and tell me your opinion. Thanks and kind regards.
Recall that the $n$-torus $T^n$ is defined as the quotient $\Bbb{R}^n/G$ where $G$ is the group of integer translations in $\Bbb{R}^n$. It can be identified with the product of $n$ circles:
$$
T^n = \underbrace{S^1 \times \ldots \times S^n}_{n \text{ times}}.
$$
We can then define a natural projection $\pi: \Bbb{R}^n \longrightarrow T^n$ given by
$$
\pi(x) = (e^{ix_1}, \ldots, e^{ix_n}), \quad x = (x_1, \ldots, x_n) \in \Bbb{R}^n.
$$
For $u = (u_1, \ldots, u_n) \in T_x\Bbb{R}^n = \Bbb{R}^n$ we have
$$
d\pi_x(u) = J(x)u = i(u_1 e^{i x_1}, \ldots, u_n e^{i x_n}).
$$
where $J(x)$ is the Jacobian matrix of $\pi$ at $x$ given by
$$
J(x) = \begin{bmatrix}
ie^{ix_1} & 0 & \cdots & 0 \\
0 & ie^{i x_2} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & i e^{i x_n}
\end{bmatrix}.
$$
We can just define
$$
\langle d\pi_x(u), d \pi_x(v) \rangle_{\pi(x)} = \langle u, v \rangle, \quad u, v \in T_x\Bbb{R}^n = \Bbb{R}^n
$$
where $\langle \cdot, \cdot \rangle$ denotes the inner product of the Euclidean space. To get a local isometry, it suffices to restrict $\pi$ to a neighborhood of $x$ such that it is a diffeomorphism.
We now show that the identity map $i: \Bbb{R}^n/G \longrightarrow T^n$ is an isometry, that is, we show that the two metrics are the same. Let $u,v \in T_pT^n$. Then
\begin{align*}
\langle u, v \rangle_{(e^{i x_1}, \ldots, e^{i x_n})} & = \sum_1^n \langle d \pi_j (u), d \pi_j(v) \rangle_{e^{i x_j}} \\
& = \sum_1^n \langle u_j e^{i (x_j + \pi/2)}, v_j e^{i (x_j + \pi/2)} \rangle_{e^{i x_j}} \\
& = \sum_1^n u_j v_j \\
& = \langle u, v \rangle,
\end{align*}
which completes the proof.
Best Answer
If the action is by isometries then yes. If $p\in M$ then $\pi_*$ gives an isomorphism of $(T_p(G\cdot p))^\perp\subset T_p M$ with $T_{\pi(p)}(M/G)$, so $T_{\pi(p)}(M/G)$ gets from $(T_p(G\cdot p))^\perp\subset T_p M$ an inner product, and the inner product doesn't change if we replace $p$ by $g\cdot p$ (for any $g\in G$).
If $G$ doesn't act by isometries then the answer depends on what you consider natural. If e.g. $G$ is compact, we can do the above-described procedure to get an inner product $(,)_p$ on $T_{\pi(p)}(M/G)$, which now depends on the choice of $p$, and then average $(,)_p$ using the Haar measure on $G$ (i.e. define $ (u,v)=\int_G (u,v)_{g\cdot p}dg$ ).