Complexity of counting graph endomorphisms
Counting endomorphisms or fixed-point-free endomorphisms is complete for $\mathsf{FP}^{\mathsf{\# P}}$: given a connected graph $G$, consider the graph $G'$ which is the disjoint union of $G$ and a triangle. Then $|\text{End}(G')| = (|\text{End}(G)| + \# 3COL(G))(\#\{\text{triangles in } G\} + 3^3)$, so $\# 3COL$ can be computed using two endomorphism counts (and by a general result, even just one suffices) and some poly-time post-processing. Note that the number of triangles can be counted in cubic (or even matrix multiplication) time. The same equation holds for fixed-point free endomorphisms, since the 3-colorings and triangles are fixed-point-free endomorphisms of $G'$. 

If you would like $G'$ to be connected, you can do as follows. First note that counting vertex-colored graph endomorphisms (where vertices of color $c$ can only be mapped to other vertices of color $c$) is equivalent to counting graph endomorphisms, as follows. Let the colors be $\{1,...,C\}$. For each vertex $v$ of color $c$, add in a new disjoint odd cycle $C_v$ of size at least $n + 2c$ ($n=|V(G)|$), and connect one vertex of $C_v$ to $v$. Every endomorphism of $G$ corresponds to $2^n$ endomorphisms of the new graph (for each cycle, you have two choices of how to map it). Note that no vertices of $G$ can map to vertices of any $C_v$, since the cycles are too large (you'd have to be able to fit one cycle inside another, which you can't for odd cycles). 

Now, to make a version of $G'$ that is connected, we start with a colored version, and then apply the above transformation. Start as before, by adding to $G$ a disjoint triangle $\Delta$. Now add a single new vertex $v_0$ that is connected to every vertex in $G \cup \Delta$. Color $v_0$ red and all other vertices blue.