EDIT: I fixed the issue pointed out by Nicholas Tholozan, thanks for sheding light on this!
This question is written as a follow-up to this one.
Both answers there are great, but my impression is there is a little more in general that one could ask about the mapping class group of $\mathbb RP^n$, i.e. the group $\Gamma := \pi_0 \text{Diff}(\mathbb RP^n)$, whose conjugacy classes are in bijective correspondence to smooth $\mathbb RP^n$-bundles over a circle (in the smooth sense).
Let $\Gamma' = \pi_0 \text{Diff}(S^n)$. We will define a homomorphism $\Gamma \to \Gamma'$ and it seems worthwhile to me to study this homomorphism. But first, let me describe what is known about $\Gamma'$. For this, note that looking at the differential of a chosen point in $S^n$ gives a fiber sequence $$\text{Diff}_{\partial}(D^n) \to \text{Diff}(S^n) \to \text{Fr}(S^n), $$ where $\text{Fr}(S^n)$ denotes the total space of the frame bundle of $S^n$. By Gram-Schmidt orthonormalization, we see that the map $O(n+1) \hookrightarrow \text{Diff}(S^n) \to \text{Fr}(S^n)$ is a homotopy equivalence, hence the long exact sequence of homotopy groups of the fiber sequence above splits into short exotic sequences. Moreover, it is well-known that for $n \geq 5$, the mapping class group $\pi_0 \text{Diff}(D^n)$ is ismorphic to $\Theta_{n+1}$, the group of homotopy $(n+1)$-spheres up to $h$-cobordism. We thus obtain a short exact sequence of groups: $$ \Theta_{n+1} \to \Gamma' \to \mathbb Z/2$$
To define the homomorphism $\Gamma \to \Gamma'$, note that conjugation by the antipodal map $\iota \colon S^n \to S^n, x \mapsto -x$ defines an action of $\mathbb Z/2$ on $\text{Diff}(S^n)$. The fixed points of this action are those diffeomorphisms of $S^n$ that induce a diffeomorphism of $\mathbb RP^n$, and thus we obtain a $2$-fold covering $p \colon \text{Diff}(S^n)^{\mathbb Z/2} \to \text{Diff}(\mathbb RP^n)$ and an inclusion $j \colon \text{Diff}(S^n)^{\mathbb Z/2} \to \text{Diff}(S^n)$. The two lifts of a given diffeomorphism $f$ of $\mathbb RP^n$ along the covering $p$ differ by $\iota$.
Now let us distinguish the cases $n$ odd/even.
If $n$ is odd, then $\text{id}_{S^n}$ and $\text{id}$ can be joined by a path within $\text{Diff}(S^n)^{\mathbb Z/2}$. Hence $p$ is non-trivial (on each component) and thus induces an isomorphism on $\pi_0$, thus $\pi_0 \text{Diff}(S^n)^{\mathbb Z/2} = \Gamma$. Hence the inclusion $j$ defines a homomorphism $\Gamma \to \Gamma'$. The composition of this homomorphism with the homomorphism $\Gamma' \to \mathbb Z/2$ described above is clearly surjective (look at reflections along hyperplanes).
If $n$ is even, the two lifts of any $f \in \text{Diff}(\mathbb RP^n)$ along $p$ differ by $\iota$, hence one is orientation-preservering and the other one orientation reversing. This proves that $p$ is a trivial covering; by always picking the orientation-preserving lift we get a section of $p$. Composed with $j$ this induces a homomorphism $\Gamma \to \Gamma'$ that yields the trivial morphism when composed with $\Gamma' \to \mathbb Z/2$.
(An aside: it seems to me that the preceeding discussion whether the homomorphism $\Gamma \to \Gamma' \to \mathbb Z/2\mathbb Z$ that we discussed is surjective or zero is what the OP of the linked question really wanted to know.)
So now two questions that immediately came to my mind and that I could not answer are:
(a) Is the homomorphism $\varphi \colon \Gamma \to \Gamma'$ injective?
(b) Is the obvious inclusion $\text{im}(2\Theta_{n+1} \hookrightarrow \Gamma') \subset \text{im}(\varphi) \cap \text{ker}(\Gamma' \to \mathbb Z/2)$ an equality?
I doubt the answer to both of these questions is affirmative, but if so this would give a nice description of $\Gamma$...
