Let a connected algebraic group $G$ (over $\mathbb C$, say) act Hamiltonianly on an algebraic symplectic variety $M$, with moment map $\Phi: M\to \mathfrak g^*$. In the example I care about, vaguely related to Slodowy slices, $G$ is unipotent.
Is $\Phi^{-1}(0)$ necessarily irreducible?
Moreover $M$ contains an irreducible Lagrangian subvariety $L$.
Is $\Phi^{-1}(0) \cap L$ necessarily irreducible?
Work of Gan and Ginzburg (https://arxiv.org/pdf/math/0409262v7.pdf) shows that one of our favorite examples, $M=T^*(\mathfrak{gl}_n\oplus \mathbb C^n)$ and $G=GL(n)$, has a 0 level of the moment map which has $n+1$ components and so is not irreducible (but for a given stability condition, all but one component is unstable).
EDIT: I'll just add that this situation can get quite bad; in the example above, at least things are equidimensional. Even this can fail, and $\Phi^{-1}(0)$ can have a larger dimension than you would expect from the GIT quotient, since only a component of smaller dimension has stable points; this happens for all quiver varieties corresponding to non-dominant weight spaces. For example, if $M=T^*\mathrm{Hom}(\mathbb C^n,\mathbb C^m)$ and $G=GL(n)$, then we have a pair of matrices $A$ and $B$ subject to $BA=0$ as the moment map condition. Obviously, the ranks of these matrices can sum to no more than $m$ or to $2n$. If $2n\leq m$, then there's a single component. If $m<2n$ (which is the non-dominant case; this is a $\mathfrak{sl}_2$ quiver variety), then there are $m-2n+1$ components, given by the closures of the sets where $A$ has rank $k$ and $B$ rank $m-k$. The dimension of this space is given by choosing a dimension $k$ subspace in $\mathbb C^m$, the surjective map of $\mathbb C^n$ to this space, and injective map from its quotient. This gives dimension $\binom mk+nm$. Thus, the largest component is when $k\approx m/2$, but the two different stability conditions are that $A$ is injective or that $B$ is surjective (so $k=n, m-n$); thus, if $n>m/2+1$, we're no longer equidimensional.
