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If H is a homeomorphism from $R^{n}$ to itself, and P and H(P) are compact polyhedra, is H(P) piecewise-linear homeomorphic to P?

I know that, in a manifold of dimension $\geq$ 5,there can exist polyhedra P and Q that are homeomorphic but not piecewise-linear homeomorphic. Can this happen if P and Q are compact subsets of $R^{n}$ and the homeomorphism maps $R^{n}$ to itself?

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Suppose that $P, Q$ are $k$-dimensional polyhedra in ${\mathbb R}^n$ which are homeomorphic but not PL homeomorphic. If $2k+2\le n$ then the homeomorphism $f: P\to Q$ extends to a homeomorphism ${\mathbb R}^n\to {\mathbb R}^n$ by H.Gluck's theorem 1.3 from his "Embeddings in the trivial range" paper (Annals of Math., 1965). Gluck's result deals with embeddings to general manifolds, so the case when the target is ${\mathbb R}^n$ may have been known earlier. – Misha Mar 23 2012 at 4:42

The answer is: Yes, this can happen. Suppose that $P, Q$ are $k$-dimensional polyhedra in ${\mathbb R}^n$ which are homeomorphic but not PL homeomorphic. If $2k+2\le n$ (and you can always increase the dimension of the ambient Euclidean space) then the homeomorphism $f:P\to Q$ extends to a homeomorphism ${\mathbb R}^n\to {\mathbb R}^n$ by theorem 1.3 from H.Gluck's "Embeddings in the trivial range" (Annals of Math., 1965). Gluck's result deals with embeddings to general manifolds, so the case when the target is ${\mathbb R}^n$ may have been known earlier.