Does bicyclo[3.2.0]heptane have a plane of symmetry? Most stable form of cyclobutane is the "envelope form" and most stable
  conformation of cyclopentane is either the "envelope form" or the "twist"
  form.



In the above compound do the individual rings exist in their most stable conformation? Or is the actual situation different?
What is the most stable conformation of the above compound? And does it have a plane of symmetry?
It turns out that bicyclo[3.2.0]heptane does have a plane of symmetry (courtesy PubChem):





According to The Molecular Structures of Bicyclo[3.2.0]heptane and $\Delta^6$-Bicyclo[3.2.0]heptene. An Electron-diffraction Study of Gaseous $\ce{C7H12}$ and Molecular Mechanics Calculations on $\ce{C7H12}$ and $\ce{C7H10}$. Robert Glen, Grete Gundersen, Peter Murray-Rust, and David W. H. Rankin, here are the bond angles in degrees:

The link explains the experimental procedure in deduction of the actual structure.There are two possible stable structures/forms for the molecule namely endo and exo.The endo form is supposed to be more favourable than the exo form.







From here, we can conclude Cs symmetry.