Conductance and diameter in regular graphs Given an undirected, regular graph $G=(V,E)$, what is the relationship between its diameter - defined as the largest distance between two nodes - and its conductance, defined as $$\min_{S \subset V} ~\frac{e(S,S^c)}{\min(|S|,|S^c|)},$$ where $e(S,S^c)$ is the number of edges crossing between $S$ and $S^c$. 

More concretely suppose I know the diameter is at least (or at most) $D$. What does this tell me about the conductance, if anything? And, conversely, suppose I know the conductance is at most (or at least) $\alpha$. What does this tell me about the diameter, if anything?
As Hsieh notes, your definition of conductance is off from the one I know by a factor of $d$, where $d$ is the degree of the regular graph. This is also known as edge expansion for regular graphs.

A relationship between edge expansion and diameter is quite easy to show. Intuitively, an expander is "like" a complete graph, so all vertices are "close" to each other. More formally, let 

$$
\min_{S \subseteq V}\ { \frac{ e(S, S^c) }{ d \cdot \min\{|S|, |S^c|\} }} \geq \alpha
$$

Take any set of vertices $S$ with $|S| \leq |V|/2$. There are at least $\alpha d |S|$ edges coming out of $S$ and since $G$ is $d$-regular, the neighborhood of $S$ (including $S$ itself) is of size at least $(1+\alpha)|S|$. Applying this claim inductively, starting from $S = \{u\}$ for any vertex $u$, we see that for some $t = O(\log_{1 + \alpha } |V|)$, $u$'s $t$-hop neighborhood has size at least $|V|/2$. Therefore, the $t+1$-hop neighborhood of any vertex $v$ has to intersect the $t$-hop neighborhood of $u$, or the graph would have more than $|V|$ vertices, a contradiction. So you have 

$$ 
D = O\left(\frac{\log |V|}{\log (1 + \alpha)}\right)
$$

Of course, it also follows that having a lower bound on the diameter implies an upper bound on edge expansion.

I don't think small diameter implies conductance. If you don't insist on regular graphs (and use Hsieh's definition), then two complete graphs connected by a single edge provides a counterexample.