Let $E_i$ be a $\mathbb R$-Banach space and $\Omega_i\subseteq E_i$. $g:\Omega_1\to E_2$ is called $C^1$-differentiable if $$g=\left.\tilde g\right|_{\Omega_1}\tag1$$ for some $E_2$-valued $C^1$-differentiable function $\tilde g$ on an open neighborhood of $\Omega_1$. Moreover, $f:\Omega_1\to\Omega_2$ is called $C^1$-diffeomorphism if
- $f$ is a homeomorphism from $\Omega_1$ onto $\Omega_2$;
- $f$ and $f^{-1}$ are $C^1$-differentiable.
So, if $f:\Omega_1\to\Omega_2$ is a $C^1$-diffeomorphism, then $$f=\left.\tilde f\right|_{\Omega_1}\tag2$$ for some $E_2$-valued $C^1$-differentiable function $\tilde f$ on an open neighborhood $N_1$ of $\Omega_1$ and $$f^{-1}=\left.\tilde g\right|_{\Omega_2}\tag3$$ for some $E_2$-valued $C^1$-differentiable function $\tilde g$ on an open neighborhood $N_2$ of $\Omega_2$.
Is it possible to choose $N_1,\tilde f,N_2,\tilde g$ so that $\tilde f$ is a $C^1$-diffeomorphism (in the ordinary sense) from $N_1$ onto $N_2$ and $\tilde g=\tilde f^{-1}$?
I think you need more assumptions. To be explicit, consider the $x$-axis in $\mathbb{R}^2$ and $\mathbb{R}^3$. These are diffeomorphic, but clearly don’t have any diffeomorphic neighborhoods.