Explicit isomorphism $\pi_{n+1}(\mathbb{RP}^n) \cong \pi_1(\mathbb{RP}^{n-1})$ 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


*

*valid all $n$

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

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

*induced by a map (optional)
Does $\lambda_n: \mathbb{RP}^{n-1} \to \Omega^n\mathbb{RP}^n$ satisfies these conditions?
 A: Let $n>2$. You have a map $\lambda_n:\mathbb{R}P^{n-1}\to\Omega^n S^n$ defined using reflection maps. This is the map that leads to the Kahn-Priddy theorem. This map extends to an $n$-fold loop map 
$$\lambda_n:\Omega^n\Sigma^n\mathbb{R}P^{n-1}\to\Omega^n S^n$$ 
which according to Kahn-Priddy Theorem induces an epimorphism on ${_2\pi_i}$ for $0<i<n-1$. The inclusion map $\mathbb{R}P^{n-1}\to \Omega^n\Sigma^n\mathbb{R}P^{n-1}$ induces an isomorphism on ${\pi_1}$. Now, from knowing that $\pi_1\Omega^nS^n\simeq\pi_1\mathbb{R}P^{n-1}\simeq\mathbb{Z}/2$ you can deduce that the composition 
$$\mathbb{R}P^{n-1}\to\Omega^n\Sigma^n\mathbb{R}P^{n-1}\to\Omega^nS^n$$
induces an isomorphism on ${_2\pi_1}$ which gives the desired isomorphism on ${\pi_1}$. Note that the geometric description of $\lambda_n$ is quit explicit.
ADDED Since the title of question is about $\mathbb{R}P^{n-1}$ and $\Omega^n\mathbb{R}P^n$, then it would suffice to compose the above composition with the $n$-loop of the covering map $S^n\to\mathbb{R}P^n$ which yields a map
$$\mathbb{R}P^{n-1}\to\Omega^n\mathbb{R}P^n$$
inducing the desired isomorphism. 
