$\mathbb{R}^k$ and $\mathbb{R}^k$ are trivially diffeomorphic. Is this claim correct?
If so, is it because identity is the diffeomorphism?
 A: Yes, the identity map is a diffeomorphism. The reason is that it is smooth and its inverse (the identity map again!) is smooth. Finally, its bijective for obvious reasons.
The following exercise provides more interesting examples of diffeomorphisms from $\mathbb{R}^{k}$ to $\mathbb{R}^k$.
Exercise 1: Let $T:\mathbb{R}^{k}\to \mathbb{R}^{k}$ be the linear operator defined by the invertible $k\times k$ matrix $A$ by the rule $T(x)=Ax$ for all $x\in \mathbb{R}^{k}$. Prove that:
(a) $T$ is bijective
(b) $T$ is smooth and its derivative at all points of $\mathbb{R}^{k}$ is $A$
(c) $T^{-1}$ is smooth and its derivative at all points of $\mathbb{R}^{k}$ is $A^{-1}$
Therefore, $T:\mathbb{R}^{k}\to \mathbb{R}^{k}$ is a diffeomorphism!
Exercise 2: Prove or give a counterexample: every diffeomorphism from $\mathbb{R}^{k}$ to itself is of the form given by Exercise 1.
Exercise 3: Prove that if $m\neq n$, then $\mathbb{R}^{m}$ is not diffeomorphic to $\mathbb{R}^{n}$. (Hint: use the implicit function theorem.)
I hope this helps!
