Let $g$ be a Riemannian metric $\mathbb{RP}^n$, with no isometries except the identity.
Let $\pi:\mathbb{S}^n \to \mathbb{RP}^n$ be the natural projection, and consider the pullback metric $\pi^*g$ on $\mathbb{S}^n$.
Is it true that the only isometries of $(\mathbb{S}^n,\pi^*g)$ are the identity and the antipodal map?
(To see that $f(x)=-x$ is an isometry, note that $\, f^*(\pi^*g)=(\pi \circ f)^*g=\pi^*g. $
(Motivation: I am trying to construct metrics with discrete set of isometries, and this seemed like a natural way).
Comment: If we can prove every isometry $\phi:(\mathbb{S}^n,\pi^*g) \to (\mathbb{S}^n,\pi^*g)$ can be projected to $\mathbb{RP}^n$ then we are done.
Specifically, suppose that $\phi(x)=\phi(-x)$ or $\phi(x)=-\phi(-x)$. Define $\tilde \phi:\mathbb{S}^n \to \mathbb{RP}^n$ by $$\tilde \phi=\pi \circ \phi.$$
Our assumption on $\phi$ implies $\tilde \phi$ can be projected to the quotient, i.e we have a map $\psi:\mathbb{RP}^n \to \mathbb{RP}^n$
where $\psi \circ \pi =\tilde \phi=\pi \circ \phi$. Then
$$ \pi^*\psi^*g=(\psi \circ \pi)^*g=(\pi \circ \phi)^*g=\phi^* \pi^*g=\pi^*g,$$ so $\psi^*g=g$, i.e $\psi$ is an isometry, so $\psi=\text{Id} \Rightarrow \pi =\pi \circ \phi$, so $phi$ is the identity or the antipodal map.
This is not an answer but a long comment.
I find it very odd that you regard pull-back of metrics on $RP^n$ as a natural way to construct metrics with discrete symmetry groups on $S^n$. A much simpler way is to multiply the standard metric on $S^n$ by some smooth positive scalar function $u$. For most choices of $u$, the resulting metric will have no nontrivial self-isometries.
As for your question: The isometry group of a Riemannian metric $h=\pi^* g$ on $S^n$ is a compact subgroup $G$ of $Diff(S^n)$, containing the antipodal involution $t: x\to -x$. If $t$ belongs to the center of $G$ then every $\phi\in G$ descends to an isometry of $(RP^n,g)$.
Conversely, if $G$ is a compact subgroup of $Diff(S^n)$, there exists a $G$-invariant Riemannian metric $h$ on $S^n$. Suppose that $t\in G$ is an involution acting freely on $S^n$. Then $S^n/<t>$ is a homotopy $RP^n$, see [1]. Assume that $t$ is non-exotic, i.e. this quotient is diffeomorphic to $RP^n$. If $t$ has trivial centralizer in $G$, then none of the elements of $G$ projects to a nontrivial diffeomorphism of $RP^n$. If $t$ has nontrivial centralizer $C_G(t)$ in $G$ (i.e the centralizer is strictly larger than $<t>$), then replace $G$ with $C_t(G)$, which is again a compact Lie group.
This, your question is equivalent to:
Question. Suppose that $G$ is a compact subgroup of $Diff(S^n)$ containing a fixed-point free non-exotic involution $t$. Does $t$ belong to the center of $G$?
(The positive answer to this question is equivalent to the positive answer to your question.)
(a) The answer is positive if $G$ is topologically linearizable, i.e. is topologically conjugate to a subgroup $H$ of $O(n)$: Under this conjugation, $t$ corresponds to $-I$ (the only orthogonal transformation having no fixed unit vectors), hence, central in $H$, hence, $t$ is central in $G$.
(b) It is known that all compact subgroups of $Diff(S^n)$, $n\le 4$, are smoothly linearizable. Hence, the answer is positive in this case as well.
(c) There is a (nontrivial) paper by Milnor where he analyses the case of finite groups acting freely on $S^n$. In this case, the answer to the question is again positive.
(d) If $G$ is a compact Lie group of positive dimension and $t\in G$ is an involution then $C_G(t)\ne \langle t\rangle$. Hence, the question is only genuinely interesting for finite groups $G$. In particular:
Answering your question completely would require a nontrivial analysis of involutions in non-linearizable compact subgroups of $Diff(S^n)$ which is much harder than finding metrics on $S^n$ with discrete symmetry group (see item 1).
If you really care about the problem, a place to start is the book
[1] Lopez de Medrano, "Involutions on Manifolds",
which is mostly about fixed-point free involutions acting on spheres.