Show that $\bar{F} : \partial \mathbb{H^n} = \mathbb{\mathbb{R^{n-1}}} \rightarrow \partial \mathbb{H^n} $ is a diffeomorphism

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Let $F : \mathbb{H^n} \rightarrow\mathbb{H^n} $ ($\mathbb{H^n}$ = upper half space) is an orientation preserving diffeomorphism.

Then how do I show that $F$ induces a diffeomorphism $\bar{F} : \partial \mathbb{H^n} = \mathbb{\mathbb{R^{n-1}}} \rightarrow \partial \mathbb{H^n} $ on the boundary which is also orientation preserving ?

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Hints: You know what map you're working on. The restriction of a smooth map to a smoothly-embedded submanifold is still smooth. Since the inverse of $G$ is the restriction of the inverse of $F$, you should be well on your way.

Can you show that $F(\partial H^n) \subset \partial H^n$? (You need that to define $G$ at all!)

Orientation is the only slightly tricky part. Suppose that $t$ is tangent to $\partial H^n$ at $x$. Can you show that $DF(x)(t)$ is tangent to $\partial H^n$ at $F(x)$? Do that $n-1$ times, and then take as an $n$th basis vector $u_n = (0,0,\ldots, 1)$. What can you say about the last coordinate of $DF(x)( u_n )$?