Riemannian metric in the projective space

You are defining $\mathbb{P}^n = \mathbb{S}^n / p\sim A(p)$. Let's assume that you have verified to your satisfaction that this action is "nice" enough to produce a smooth quotient manifold.

Let's consider any point $p\in\mathbb{P}^n$. The only thing we know about $p$ is that it's an equivalence class $p = [x] = \{x_1, x_2\}$ which are related by $x_i = A(x_j)$. The tangent spaces $T_{x_1}\mathbb{S}^n$ and $T_{x_2}\mathbb{S}^n$ are related by $dA$, and via this identification we can consider either of them a model for $T_p\mathbb{P}^n$. So if we want to put a metric on $T_p\mathbb{P}^n$, there's really only one thing we can do: pick the metric on one of the tangent spaces $T_{x_i}\mathbb{S}^n$.

Now we worry! We've made a choice of identification. Does it agree with all the other possible choices we could have made? In fact, yes, precisely because $A$ is an isometry. The metric on $T_{x_j}\mathbb{S}^n$ pulled back by $dA$ is exactly what we would have picked if we had chosen $T_{x_j}\mathbb{S}^n$ to define our metric.

This strategy works whenever $Y = X/\Gamma$, where $X$ is a Riemannian manifold and $\Gamma$ is a discrete group acting freely on $X$ by isometries. Then the quotient map is a covering and the covering neighborhoods are all isometric, so the choice of metric is natural.

The general case $Y = X/G$, $X$ Riemannian and $G$ is any group acting by isometries, is more complicated.

Note first that the projective space, $\mathbb{P}^n$, is the quotient space, $$\mathbb{P}^n=\mathbb{S}^n/ \Gamma$$ where, $\Gamma= \{A,Id\}$. We obtain an atlas for $\mathbb{P}^n$ simply choosing a neighborhoods $U$ of $\mathbb{S}^n$ such that $\pi|U$ is injective, where $\pi$ is the canonical projection. In this way, $\pi|U=A|U.$

Now fix $p$ in $\mathbb{P}^n$ and let $u,v\in T_p \mathbb{P}^n$. Define,

$$<u,v>_p=<d(\pi|U)^{-1}u,d(\pi|U)^{-1}v>_{\pi^{-1}(p)}.$$

with this metric the canonical projection is an local isometry as defined by what has already been discussed above.