Look at the $d$-dimensional hypercube and truncate it. This means one replaces each vertex by a cycle (of length $d$) in such a way the the new graph is 3-regular.
Question 1: What is the asymptotic as $d \to \infty$ of the Cheeger (or isoperimetric or conductance) constant of $TQ_d$?
By asymptotic I mean $\Theta$ (actually $\Omega$ is probably enough; see below). I am using the convention: $$ h(G) = \displaystyle \min_{|X|\leq |G|/2} \frac{|\partial X|}{|X|} $$ where $|\partial X|$ is the set of edges between $X$ and $X^\mathsf{c}$. But since the graph is 3-regular, the various definitions are equivalent up to multiplicative constants (i.e. they are $\Theta$ of each other).
If you cut the hypercube along an hyperplane (parallel to the coordinate planes), then one gets that $h(TQ_d) \leq \frac{1}{d}$. And I am relatively sure there should be a lower bound (i.e. $h(TQ_d)= \Omega(\frac{1}{d})$ but could not find it in the literature.
Also, there is a lower bound on $h$ using the spectral gap $\lambda$ ($h \geq K\lambda$ for some $K$). But $\lambda$ is $O(\frac{1}{d^2})$. Note that the upper bound $h \leq K' \sqrt{\lambda}$ implies $h = O(\frac{1}{d})$ and is coherent with the above.
Question 2: does it matter what kind graph one uses to replace the vertices (instead of a cycle)?
There are two extreme cases (where the replacement graph has either the best or the worst Cheeger constant):
one could either replace a vertex in $Q_d$ by a sequence of expanders $X_d$ (with $|X_d| = d$ and degree bounded by, say, 10)
one could replace a vertex by a line (remove one edge from each cycle)
In the expander case, it's (hopefully) easy to see that the best "cuts" are those from the hypercube.
