Neighborhood theorem for conical Lagrangian Let  $(M,\omega)$ be a compact $2n$ dimensional symplectic manifold and $T$ be a compact smooth $(n-1)$ dimensional manifold.
Let $CT$ be the cone over $T$, i.e. $CT = T\times [0,1] / \sim $ where $\sim$ is an equivalence relation $(x,1)\sim(y,1)$. Clearly $CT$ is a smooth manifold away from the point $[(y,1)]$.
A conical Lagrangian $C \subset M$ over $T$ is a homeomorphism $\phi:CT\to M$ whose image is $C$ and which is a smooth Lagrangian embedding away from the conical singular point.
Is there a softer version of Weinstein Theorem which says that if $C’$ is a conical Lagrangian over $T$ in a different symplectic manifold $M’$ , there are neighborhoods of $C$ and $C’$ which are symplectomorphic?
 A: The answer is 'no', even in the simplest nontrivial case $n=2$, if you assume that the symplectomorphism is to take $C$ to $C'$.  (If you don't assume that the symplectomorphism is supposed to take $C$ to $C'$, then I think it's 'yes'.)
In the case $n=2$, you can take $T\simeq S^1$, so $CT$ is topologically a disk $\Delta = \{z\in\mathbb{C}\,|\,|z|<1\}$.  Let $M$ be a complex algebraic surface with holomorphically trivial canonical bundle, say $M = \mathbb{C}^2/\Lambda$ where $\Lambda\subset\mathbb{C}^2$ is a suitable lattice, and let your symplectic structure be given by the real part of the nonvanishing holomorphic $2$-form.  Then any injective holomorphic map $\phi:\Delta\to M$ whose differential only vanishes at $z=0$ will give you a conical Lagrangian $\phi(\Delta)\subset M$, but the various different types of singularities of $\phi$ at $z=0$ will give you conical Lagrangians that are not even topologically equivalent near $\phi(0)$ because their links will be topologically distinct.  For example, in local holomorphic coordinates $(w_1,w_2)$ near a point of $M$, you could have $\phi(z) = (z^p,z^q)$ where $p>q>1$ are relatively prime integers, and all of these will be distinct topologically in a neighborhood of $w_1=w_2=0$.
