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Let $h:B^{n}\to B^{n}$ be a homeomorphism of the $n$-dimensional unit ball in $\mathbb{R}^n$. If $h(x)=x$ for $x\in S^{n-1}$, then it is relatively simple to show that $h$ is isotopic to the identity on the full $B^{n}$: I can construct the isotopy as follows: $$ h_{t}(x):= \left\{ \begin{array} xx & \mbox{if } t\leq |x| \leq 1 \\ t\cdot h\left(\dfrac{x}{t} \right) & \mbox{if }0\leq |x| \leq t \end{array} \right. $$

My question is: if I just assume that $h$ is isotopic to the identity on $S^{n-1}$, does it still follow that $h$ is isotopic to the identity on the full ball $B^{n}$?

I want to use this result to prove that if I have two homeomorphisms $h,k: B^{n}\to B^{n}$ such that their restrictions to $S^{n-1}$ are isotopic, then they are isotopic on $B^{n}$. Any hints or suggestions would be very much appreciated!

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    Yes - you "cone off" the isotopy in the same way. Your formula can be easily modified so that your isotopy is an isotopy to the cone on a homeomorphism of $S^{n-1}$, and then you just cone off the isotopy to the identity – 2017-01-09

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Here's the rough idea, coming from what Mike Miller said in the comments. We will view the ball $B^n$ as a union of concentric spheres. Let $B_+$ denote the union of the spheres of radius at least 1/2 and $B_-$ denote the union of the spheres of radius at most $1/2$ (including the origin as well.) Note that $B_+\cap B_-$ is the sphere of radius $1/2$.

We will first isotope $h$ to $h_1$, where $h_1$ copies $h$ on all spheres in $B_+$, and then "speeds" through $h$ on $B_-$.

In formulas, we use $H_1(x,t) = \begin{cases}|x| h\left(\frac{x}{|x|}\right) & \text{if } t\leq |x| \\ th\left(\frac{x}{t}\right) & \text{if } |x|\leq t \end{cases}$ where we are thinking of $t\in [1/2,1]$. Then $H_1(x,1) = h(x)$ and $H_1(x,1/2) = h_1$ is described above. In particular, $h_1$ on the sphere of radius 1/2 is just a scaled version of $h$.

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Next, we isotope $h_1$ to $h_2$ where $h_2|_{B_-} = h_1|_{B_-}$ but where $h_2|_{S^{n-1}} = Id_{S^{n-1}}$.

To do this, first pick an isotopy $F:S^{n-1}\times [0,1]\rightarrow S^{n-1}$ between $h|_{S^{n-1}}:S^{n-1}\rightarrow S^{n-1}$ and the identity with $F(x,0) = h$ and $F(x,1) = Id$. Then, to isotope $h_1$ to $h_2$, we use the formula $H_2(x,t) = \begin{cases} |x|F(x/|x|, (2|x|-1)t) & \text{if } |x|\geq 1/2\\ h_1(x) & \text{if } |x| \leq 1/2\end{cases}.$

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At this stage, $h_2$ is the identity on $S^{n-1}$, so to finish isotoping $h$ to $Id|_{B^n}$, we apply the homotopy you gave in your post.