**Obsolete answer to initial version of OP.** (and, to set the record straight: there was an error in my answer, and Mikhail Tikhomirov's was the only correct answer. The original version of the OP allowed *empty graphs as the targets in the mapping property that the OP was asking for*, and the only graph which admits a graph-homomorphism *to* the empty graph is the empty graph. Symbolically, the slice category $\mathsf{IrreflexiveSymmetricBinaryRelationsWithGraphHomomorphisms}/(\{\},\{\})$ equals the full subcategory consisting of the sole object $(\{\},\{\})$. It was *wrong* in my hasty answer to claim that the one-vertex graphs are weakly-initial: they are *not*, since they do not map into the empty graph. I will leave the answer as it is, for several reasons, yet will add a a warning.

Echoing Mikhail Tikhomirov's answer a bit: the answer to the question in the OP is evidently yes: EDIT: (what comes next is wrong) the (still: proper) class of one-vertex graphs (i.e., isomorphic copies of $K^1$) is the class of weakly initial objects of the category implicit in the OP, ENDOFWRONGSTATEMENT and [the following **was** true as long as the OP still allowed non-connected graphs; the empty graph is not connected] the empty graph is its unique initial object.

weakly initial object. $\endgroup$ – Peter Heinig Sep 9 '17 at 10:07