Is there a Fraisse limit whose automorphism group contains dense but not generic automorphisms? It is well known that $\mathsf{Aut}(\mathbb{Q},<)$ has generic automorphisms (i.e., a comeagre conjugacy class under the diagonal action) but does not admit ample generics. The automorphism group $\mathsf{Aut}(\omega,E)$ of the structure $(\omega,E)$ where $nEm \iff n = m \mod{2}$ does not even have a dense conjugacy class. I am looking for an example 'in the midst' of the previous two, i.e., an automorphism group of a Fraisse limit that has a dense conjugacy class but does not have a generic automorphism. 
I know that there are examples of Polish groups that have a dense conjugacy but do not have a generic automorphism. However, it is unknown to me or impossible to construct these examples as automorphism groups of Fraisse limits.
The first example that I know of is $\mathsf{Aut}(X,\mu)$ where $(X,\mu)$ is the standard probability space, i.e., a standard Borel space $X$ with a probability measure $\mu$. We cannot express $(X,\mu)$ as a Fraisse limit since this space is not countable. Another example of a Polish group that has dense automorphisms is $U(H)$, that is, the unitary group of the infinite dimensional Hilbert space. However, the last example is not even constructed as an automorphism group.
Is there an easy example of a Fraisse limit whose automorphism group has a dense conjugacy class but not a comeagre one?
 A: I assume you're familiar with the paper Turbulence, amalgamation, and generic automorphisms of homogeneous structures by Kechris and Rosendal. If not, you should have a look, since it's about exactly these issues.
In the language of that paper, we take a Fraïssé class $K$ and expand it to a class $K_p$ consisting of pairs $(A,\varphi)$, where $A\in K$ and $\varphi$ is a partial automorphism of $A$ (i.e. an isomorphism between two substructures of $A$). Then an embedding $(A,\varphi)\to (B,\psi)$ is an embedding $A\to B$ such that $\psi$ extends the image of $\varphi$. If $M$ is the Fraïssé limit of $K$, Kechris and Rosendal show that $\mathrm{Aut}(M)$ has a dense conjugacy class if and only if $K_p$ has the joint embedding property (JEP), and $\mathrm{Aut}(M)$ has a comeager conjugacy class if and only if $K_p$ has the weak amalgamation property (WAP) and the JEP. 
For a category of structures and embeddings, the WAP is the assertion that every $A$, there is an embedding $f\colon A\to B$, such that for any embeddings $g_1\colon B\to C_1$ and $g_2\colon B\to C_2$, there are embeddings $h_1\colon C_1\to D$ and $h_2\colon C_2\to D$, such that $h_1\circ g_1\circ f = h_2\circ g_2\circ f$. 
So to answer your question, we just need an example of a Fraïssé class $K$ such that $K_p$ has the JEP but not the WAP. (Edit: I've realized that there's a much simpler example, so I've replaced my original example).
Let $K$ be the class of finite ordered graphs in the language $\{\leq,R\}$. Then $K$ is a Fraïssé class. Its Fraïssé limit is the ordered random graph. 
$K_p$ has the JEP: Given $(A,\varphi)$ and $(B,\psi)$, embed them in $(C,\theta)$, where $C$ is the disjoint union of $A$ and $B$, $\theta$ is the union of $\varphi$ and $\psi$, all the elements of $B$ are greater than all the elements of $A$ in the order, and no new edge relations hold.
But $K_p$ does not have the WAP: Let $(A,\varphi)$ be any structure in $K_p$ such that there is some $a\in A$ such that $\varphi(a)$ is defined and $\varphi(a)> a$. Suppose $(A,\varphi)$ embeds in $(B,\psi)$. We will show that $(B,\psi)$ cannot witness the WAP for $(A,\varphi)$. Note that the sequence $\dots, \psi^{-1}(a), a, \psi(a), \psi^2(a), \dots$ is strictly increasing, so, since $B$ is finite, there is some largest $m\geq 0$ such that $b = \psi^{-m}(a)$ is defined, and there is some largest $n\geq 1$ such that $e = \psi^{n}(a)$ is defined. We will define two structures $(C_1,\theta_1)$ and $(C_2,\theta_2)$ into which $(B,\psi)$ embeds. 
In each, we will add a single new element $c$ and extend $\psi$ to $\theta_i$ so that $\theta_i(e) = c$. For all $d$ such that $\psi(d)$ is defined, we set $\psi(d) < c$ if and only if $d < e$ and $\psi(d)Rc$ if and only if $dRe$. This is enough to ensure that the $\theta_i$ are partial automorphisms. Now in $C_1$ we set $bRc$, and in $C_2$ we set $\lnot bRc$. This is consistent with the assignments above, since $b$ does not have a preimage under $\psi$. We may decide the rest of the order and edge relations between $c$ and elements of $B$ arbitrarily (but consistently with the assignments above).
If $(C_1,\theta_1)$ and $(C_2,\theta_2)$ embed in some $(D,\rho)$ over $(A,\varphi)$, the images of $b = \theta_i^{-m}(a)$ and $c= \theta_i^{n+1}(a)$ must be equal, but $C_1$ and $C_2$ disagree about the truth of $bRc$, so we have a contradiction.

Two comments:


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*In fact, this argument shows that $K_p$ does not even have the local weak amalgamation property (i.e. there is no $A\in K_p$ such that the weak amalgamation property holds for the class of structures into which $A$ embeds), so the automorphism group of the ordered random graph does not even have any non-meager conjugacy classes. 

*A google search for "ordered random graph" generic automorphism brought me to this paper, where essentially the same argument is used to show that the automorphism group of the ordered rational Urysohn space has no non-meager conjugacy class. At the very end of the paper it is remarked that the results also hold for the ordered random graph.

