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.