Let $N$ and $N'$ be regular neighborhoods of a subpolyhedron $P$ in a closed PL manifold $M$, and suppose that $t$ is a free PL involution on $M$ such that each of $\partial N$, $\partial N'$ is invariant under $t$. Does there exist an equivariant PL isotopy of $M$ taking $N$ onto $N'$? 


**Edit:** *Beware that $N$ is not assumed to be $t$-invariant. The case where $N$ is $t$-invariant is of course trivial.* 


(Example: $M=S^m$ with $t$ the antipodal involution, $P$ is a point, $N$ is a hemisphere and $N'$ something fancier.)


I also do not know the answer to the similar question in the smooth category, where $N$ and $N'$ are tubular neighborhoods of a submanifold.


It does not seem like usual arguments about regular neighborhoods work to prove this, but also I don't immediately see a counterexample. 


(As a side remark, I'm ultimately interested in relative regular neighborhoods and a non-free involution whose fixed point set is forced to lie in $\partial N$ by the choice of the relativization.)



uniqueness of regular/tubular neighborhood with equivariant boundary
I learned from Piotr Akhmetiev that $S^6$ contains two smoothly embedded $5$-spheres invariant under the antipodal involution that are not equivariantly PL isotopic, and the reference is [Lopez de Medrano's "Involutions on Manifolds"](http://www.zentralblatt-math.org/zmath/search/?an=0214.22501). Of course they are boundaries of regular neighborhoods of a point (by the higher-dimensional Poincare conjecture) and hence also of tubular neighborhoods of a point (since there are no exotic $6$-balls). A more recent source that Akhmetiev mentioned is [Yu. Muranov's survey](http://mi.mathnet.ru/eng/tm1075).


