Unfortunately, I don't have enough reputation to comment, but there seems to be a problem with both solutions suggested so far: The graphs are bipartite, meaning that they allow a homomorphism to a single edge, which of course is a non-trivial homomorphism from the graph to itself (note that homomorphisms are not assumed to be injective).

However, here is a more complicated construction which hopefully takes care of this:

Let $H_0 = K_3$. For $i \geq 1$ let $H_i$ be a triangle free graph such that $\chi(H_i) > \chi(H_{i-1})$ and such that the only homomorphisms from $H_i$ to itself are automorphisms (there is for example a family of Kneser graphs with this property).

Enumerate the vertices of the disjoint union $\bigcup _{i \geq 0} H_i$ as $(v_j)_{j \geq 1}$ such that $v_j \in H_i$ for some $i < j$.

Now define a sequence $G_n$ as follows:

- $G_0 = H_0$.
- For $i > 0$ let $l$ be the maximum of the diameters of $G_{i-1}$ and $H_i$ and connect one vertex of $H_i$ to $v_i$ by a path of length $l$.
- Let $G$ be the direct limit of this construction.

There is no homomorphism from $K_3$ to a triangle free graph, hence $G_0$ must be fixed by every homomorphism.

There is no homomorphism from $G$ to $H$ if $\chi (H) < \chi (G)$, hence $H_i$ cannot be mapped into $G_{i-1}$ (plus the paths leading to $H_j$ for $j>i$). Since Homomorphisms never increase distances, i.e. if $f$ is a homomorphism, then $d(x,y) \geq d(f(x),f(y))$, this together with the fact that $d(H_i,H_0)$ is bounded by the diameter of $G_i$ implies that the image of $H_i$ must intersect $H_i$. Consequently $H_i$ must map to itself---if the image of $H_i$ intersected one of the paths attached to $H_i$, then this would give a homomorphism $H_i \to H_i$ which is not an automorphism.

So far we showed that each $H_i$ is fixed setwise by every self-homomorphism of $G$. Finally, the structure between the $H_i$ guarantees that every $H_i$ must be fixed pointwise: if $v_j \in H_i$, then it is the unique vertex in $H_i$ which minimises the distance to $H_j$.