Is the Lipschitz unique for close Alexandrov space other than dimension 4?

For a closed smooth $n$-manifold ($n\ne 4$), the Lipschitz structure is unique by the result of Sullivan. How about the Alexandrov spaces?

At first I was thinking it is trivially true by induction, since locally distance ball in an Alexandrov space is homeomorphic to cone over it's space of directions, which is also an Alexandrove space. But for 5-dimension and higher, the space of directions could be 4-dimensional Alexandrov spaces which admits a metric with curvature bounded from below by 1.

Now one can ask another question for manifold: "whether there exists a positively curved 4-manifold that admits two different Lipschitz structure?"

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First, I want to point out that unlike for manifolds there is no notion of a geometric Lipschitz structure for general Alexandrov spaces as they are not locally modeled on a fixed space topologically. One can still ask whether you can have two Alexandrov spaces which are homeomorphic but not bi-Lipschitz homeomorphic but that is a much more metrically rigid question. This is an open problem for $n>4$. I believe your second question is open too mostly because there are so few examples of positively curved smooth manifolds. $\mathbb{CP}^2$ and $\mathbb S^4$ are all the known orientable ones in dimension 4 and as far as I know neither is known to admit an exotic lipschitz structure.
Edit: Actually, $\mathbb {RP}^4$ may be a candidate for this. If I understand things correctly there exists an exotic smooth $\mathbb{RP}^4$ by a result of Fintushel and Stern. I'm actually somewhat confused by this because Fintushel and Stern only say that their example is homotopy equivalent to $\mathbb{RP}^4$ but don't mention homeomorphism. But I've seen people claim that Fintushel and Stern example produces a topological $\mathbb{RP}^4$. Perhaps someone in the area can clarify this. But assuming this is the case one can hope to show that their example is not bilipschitz to the standard $\mathbb{RP}^4$. I don't think this is known however. As far as I know the only tools for distinguishing lipschitz structures on 4-manifolds are due to Sullivan and Donaldson in "Quasiconformal 4-manifolds". They are Lipschitz versions of Donaldson invariants and I think that theory only works for manifolds with $b_2>1$ which is not the case here.
Note that if you could construct two homeomorphic but not bilipschitz positively curved Alexandrov manifolds $X$ and $Y$ of dimension 4 (or more generally positively curved Alexandrov spaces of dimension 4) then you'd also get a counterexample to your first question given by spherical joins $X* X$ and $Y*Y$. This is because any bilipschizt homeomorphism between $X* X$ and $Y* Y$ would need to preserve the topological strata. So it would send $X$ to $Y$ and since they are convex in $X*X$ and $Y*Y$ respectively the resulting homeomorphism $X\to Y$ would be bilipschitz with respect to the original metrics on $X$ and $Y$.
The Fintushel-Stern paper you reference sightly predates Freedman's proof of the topological $h$-cobordism theorem in dimension $4$. Hence the weaker statement. –  Adam Feb 3 '12 at 2:27
I found that in "Some new four-manifolds(Ann. of Math. (2) 104 (1976), no. 1, 61–72", Cappell and Shaneson constructed an exotic $Q^4$, which is homotpic to but not diffeomorphic to $\mathbb{RP}^4$. Then by Freedman's topological s-cobordism theorem it's homeomorphic to $\mathbb{RP}^4$. (I am not familar with Freedman's theorem though) –  John B Feb 3 '12 at 2:29
@Adam and @MG, you are quite right, thanks for the clarification. They get an h-cobordism which is an s-cobordism since the Whitehead group of $Z_2$ is trivial and that implies homeomorphism by Freedman. –  Vitali Kapovitch Feb 3 '12 at 3:06