Let $\Gamma_n$ be a family of $d$-regular finite simple graphs.
1). $\Gamma_n$ has logarithmic diameter if $diam(\Gamma_n) = O(\log |\Gamma_n|)$;
2). $\Gamma_n$ has logarithmic mixing time if
$$ \left \| v^{(k)} - \frac{1}{|\Gamma_n|} \right \|_2 \leq \frac{1}{|\Gamma_n|} $$
for some $k = O(\log |\Gamma_n|)$, where $v^{(k)}$ is the distribution of the random walk after $k$ steps;
3). $\Gamma_n$ is an expander family if it exists $\delta<1$ such that
$$ \lambda(\Gamma_n) \leq \delta \hspace{1cm} \mbox{for every } n \geq 1,$$
where $\lambda(\Gamma_n)$ is the largest non-trivial eigenvalue of the normalized adjacency matrix of $\Gamma_n$ (Markov operator).
Now, it is not hard to prove that 3) -> 2) -> 1), and there are examples where 1) -|-> 3), but I do not know about 2). One of these is the standard Cayley graph of the wreath product $C_2 \wr C_n$.
Do you know some examples where 1) -|-> 2), or 2) -|-> 3) ?
It could be better if they involve Cayley graphs.
The below answer is a sketch.
Well, 1) does not imply 2); consider a complete binary tree $T$ on $N$ vertices. A random walk where the starting point is uniformly distributed among vertices on the left side of $T$ will still be on the left side of the tree with probability $1-o(1)$ after many more than $\theta(\log N)$ steps. [You could make this graph 3-regular by adding edges between leaves that share the same grandparent in $T$ and it won't change things by much. If you really wanted a counterexample that is a family of regular Cayley graphs see the enclosed link below]
However, 2) does imply 3) though, at least for regular graphs. Here is a SKETCH: Let $G$ be the graph, $x$ be an eigenvector that corresponds to the second-largest eigenvalue of the normalized agacency matrix $A(G)$ of $G$. such that $x$ satisfies $||x||_1 = 1$. Then as the vector of all $1$s is the eigenvector corresponding to the largest eigenvalue of $A(G)$ [which is normalized to 1] and distinct eigenvectors of $A(G)$ are orthonormal to each other, it follows that $\sum_{v \in V(G)} x(v) = 0$. Then let $x^+$ be the vector where $x^+(v) = 2 x(v)$ for each $v \in V(G)$ s.t $x(v) \ge 0$, and let $x^{-}(v)$ the the vector where $x^{-}(v) = -2x(v)$ for each $v$ s.t. $x(v) <0$. Then $||x^+||_1 = ||x^-||_1 = 1$, and $x^+-x^- = 2x$. Then let consider a random walk $W_+$ where the starting distribution is picked according to $x^{+}$, and another random walk $W_-$ where the starting distribution is picked accorsing to $x^{-}$. Let $W^k_+$ be the distribution of the $k$-th step of $W_+$ and let $W^k_-$ be the distribution of the $k$-th step of $W_-$. Then $W^k_+$ and $W^k_-$ are both stochastic vectors for each positive integer $k$.
Now let us set $\delta$ to be the 2nd-largest eigenvalue of $A(G)$ [where $A(G)$ is normalized so that the largest eigenvalue of $A(G)$ is 1]. Then for each positive integer $k$:
$$||A^k(G)(2x)||_1 = \delta^k||2x||_1 = ||W^k_+ - W^k_-||_1$$
If the 2nd-largest eigenvalue $\delta$ of $A(G)$ isn't bounded away from 1, then $||W^k_+ - W^k_-||_1$ won't converge to 0 fast enough i.e., be sufficiently small for $k = \theta(\log n)$ where $n \doteq |V(G)|$. This implies that either $W^k_+$ or $W^k_-$ [or both] doesn't converge fast enough to the uniform distribution with $k$. This implies that there is a starting distribution $W$ i.e., $W \in \{W_+,W_-\}$ such that a random walk on $G$ prescribed by $W$ doesn't converge fast enough to uniform.
So to conclude: Fast mixing time for any starting distribution implies a second eigenvalue bounded away from 1 i.e., expansion.
Getting back to Cayley graphs, there are Cayley graphs of bounded degree and logarithmic diameter that are NOT expanders. This was however a recent research question that was answered, and the below contains a link to the paper:
Expander graphs and logarithmic diameter
So as there exist bounded-degree Cayley graphs of logarithmic diameter that satisfy 1) but not 3), and as we have shown above that 2) implies 3), it follows that there exist bounded-degree Cayley graphs of logarithmic diameter that satisfy 1) but not 2).