Strongly rigid regular graphs A simple, undirected graph $G = (V,E)$ is said to be strongly rigid if the identity is the only graph endomorphism.
For which positive integers $k>2$ is there a strongly rigid $k$-regular graph?
 A: Let $X(\mathcal{S)}$ be the block graph of a Steiner triple system $\mathcal{S}$ on $v$ points. The triple system consists of $b=v(v-1)/6$ triples from a set $V$ of size $v$ such that each pair of points from $V$ lies in exactly one triple. Necessarily $v\equiv1,3$ mod 6 and, if this condition holds, triple systems on $v$ points exist. The block graph has the triples of $\mathcal{S}$ as it
vertices, two triples are adjacent if they have exactly one point in common. The block graph is strongly regular.
A coclique in $X(\mathcal{S})$ is given by a set of pairwise disjoint triple, whence
$\alpha(X(\mathcal{S})) =\lfloor v/3\rfloor$. If $v>15$, the cliques of maximum size come from the triple containing a given point, and so 
$\omega(X(\mathcal{S}))=(v-1)/2$. So we see that, if $v\equiv1$ mod 6, then 
$\chi(X(\mathcal{S})) > \omega(X(\mathcal{S}))$.
Now in "Cores of geometric graphs" (arXiv:0806.1300v1), Gordon Royle and I prove that every endomorphism of the block graph of a Steiner triple system is either an automorphism, or is a homomorphism to a maximum clique. It follows that if $v\cong1$ mod 6, the block  graph of a Steiner triple system has no non-identity endomorphism. 
Finally Babai proved that almost all Steiner triple systems are asymmetric, whence it follows that almost all Steiner triple systems on $v\equiv1$ mod 6 points have no non-identity automorphism. (See L. Babai "Almost all Steiner triple systems are asymmetric" in Topics on Steiner systems. Ann. Discrete Math. 7 (1980), 37–39.) When $v>15$ all cliques of maximum size come from points of $V$ (exercise), when $v>15$ the automorphism group of a triple system and its block graph
are isomorphic.
So we have lots of strongly rigid regular graphs.
In "Homomorphisms of strongly regular graphs" (arXiv:1601.00969), David Roberson proves that the core of a strongly regular graph is either the graph itself, or is a complete graph. Hence any strongly regular graph with $\chi>\omega$ must be a core and, if the graph is asymmetric, it will not admit a non-trivial endomorphism.
I suspect that almost all Latin square graphs on a given order are strongly rigid.
There is a second way to potentially produce more examples. In a book somewhere, Gordon Royle and I prove that a triangle-free graph with diameter two and no "twinned vertices" is a core. It follows that is $X$ is connected, triangle-free and asymmetric, it is strongly rigid. Unfortunately no examples come to mind just now.
Finally none of this helps in finding finite cubic graphs with only trivial endomorphisms.
A: An infinite family of (strongly?) rigid 3-regular finite graphs can be constructed using the following graph $RC_1$ called the rigid connector:

A graph composed of the chain consisting of $n$ rigid connectors will be denoted by $RC_n$:

Finally, the required rigid 3-regular graph consists of 3 parallel chains $RC_k$, $RC_n$, $RC_m$ for pairwise distinct numbers $k,n,m$:
 
It is (more-or-less) clear that this graph is rigid. 

Is it strongly rigid, too?

