From covering space theory we know that $\pi_{n+1}(\mathbb{RP}^n) \cong \pi_{n+1}(\mathbb{S}^n)$.

From wikipedia I can notice that $\pi_{n+1}(\mathbb{S}^n) \cong \pi_1(\mathbb{RP}^{n-1})$.*

My question is: is there an explicit isomorphism $\pi_{n+1}(\mathbb{RP}^n) \cong \pi_1(\mathbb{RP}^{n-1})$? My question is motivated by the fact that $\mathbb{RP}^n \cong \mathbb{R}^n \cup \mathbb{RP}^{n-1}$ and $I^{n+1} = I^n \times I$ with $\mathring{I^n} \cong \mathbb{R}^n$.

*I "know" the standard calculation of $\pi_{n+1}(\mathbb{S}^n)$, for example via Pontryagin construction or $J$-homomorphism. I was wondering if it's possible to compute it in the way I stated.

(I posted this originally on math.stackexchange)

EDIT

I not accepted user51223's answer yet because since the groups are isomorphic for every $n$ (and not only at 2), I find artificial to distinguish the two cases. Moreover, I was looking for an isomorphism whose proof does not involve a previous knowledge of the isomorphism type of one of the groups. To be honest, I was looking for an isomorphism

1. valid all $n$

2. valid for the whole group (not only at 2)

3. which does not involve the knowledge neither of $\pi_{n+1}(\mathbb{RP}^n)$ nor of $\pi_1(\mathbb{RP}^{n-1})$

4. induced by a map (optional)

Does $\lambda_n: \mathbb{RP}^{n-1} \to \Omega^n\mathbb{RP}^n$ satisfies these conditions?