Does the quotient manifold inherit the Riemannian structure?

Let $(M, g)$ be a Riemannian manifold, and let $G$ be a group acting freely and properly on $M$. From differential geometry we know that the quotient set $M/G$, i.e., the set of the orbits, is a differential manifold and the quotient map $\pi:M \to M/G$ is a submersion.

My question is the following: there is a natural/general way for transferring the metric $g$ to $M/G$, in the two cases: (i) G acts isometrically or (ii) not.

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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 $(u,v)=\int_G (u,v)_{g\cdot p}dg$ ).