I've found a necessary and sufficient characterization for
when a relation $\Phi$ is nontrivially selffulfilling, in
the theorem below.
(As Aaron pointed out, every $\Phi$ is realized trivially
in the graph with no vertices and also in the graph with
one vertex, so by nontrivially selffullfilling, let us
insist that the graph have at least two vertices. Let us
speak of undirected graphs with no loops, so that we adopt
Aaron's correction that a binary relation $\Phi$ on
$\mathbb{N}$ is selffulfilling in graph $\langle
V,R\rangle$ just in case $x\mathrel{R} y\iff
\Phi(d(x),d(y))$ and $x\neq y$.)
Theorem. A symmetric binary
relation $\Phi$ on $\mathbb{N}$ is nontrivially
selffulfilling in some graph if and only if
$\neg\Phi(0,0)$ or $\Phi$ is not a subrelation of
$\{0\}\times\mathbb{N}\cup\mathbb{N}\times\{0\}$; that
is, if and only if $\neg\Phi(0,0)$ or it has $\Phi(n,k)$
for some $n,k\neq 0$.
Proof. If $\Phi(n,n)$ holds for some $n\neq 0$, then $\Phi$
is selffulfilling in the complete graph on $n+1$ vertices.
All vertices have degree $n$, and so only $d(x)=n$ arises
in this graph, and so only $\Phi(n,n)$ is relevant when
checking the selffulfilling property.
Otherwise, if $\Phi(n,k)$ holds for some $n\neq 0\neq k$,
but $\Phi(n,n)$ and $\Phi(k,k)$ both fail, then $\Phi$ is
selffulfilling in the bipartite graph $B(n,k)$, having $n$
nodes on the left connected to each of $k$ nodes on the
right. In this graph, every vertex has degree either $k$ or $n$, and all such vertices are connected by edges, fulfilling $\Phi$.
If $\neg\Phi(0,0)$, then $\Phi$ is selffulling in the
graph on any number of vertices, but with no edges. Again,
every vertex in this graph has $d(x)=0$, and so the only
relevant part of $\Phi$ is $\Phi(0,0)$, which fails, and
none of them are connected, as required.
Otherwise, $\Phi$ only relates numbers to $0$ and
$\Phi(0,0)$ holds. No such $\Phi$ can be selffullfilling
in a graph with at least two nodes $x\neq y$, since if
there are no edges in the graph, then $d(x)=d(y)=0$, and so
they would have to be connected because of $\Phi(0,0)$,
contradiction. And if there is an edge in the graph between
some $x$ and $y$, then $d(x)$ and $d(y)$ have some value
for which $\Phi(d(x),d(y))$, but by assumption one of those
values must be $0$, contradicting the fact that there is an
edge. QED
Note that the graphs arising in the proof have vertices only of very few degrees, which makes the selffulfilling property easier.
Here is the example I gave earlier, which does have vertices of every nonzero degree.
The relation $d(y)d(x)=1$ seems to be selffulfilling in the following graph, where each node is labeled with its degree. One can construct the graph in levels, where at each level, the degree increases by $1$, and the number of nodes on the next level is determined by the selffulfilling requirement. Each node on each level is connected to all nodes on any prior or next level.
....... etc.
11 11 11 11 11 (each 11 is connected to five 10s and six 12s)
10 10 10 10 10 (each 10 is connected to five 9s and five 11s)
9 9 9 9 9 (each 9 is connected to four 8s and five 10s)
8 8 8 8 (each 8 is connected to three 7s and five 9s)
7 7 7 (each 7 is connected to three 6s and four 8s)
6 6 6 (each 6 is connected to three 5s and three 7s)
5 5 5 (each 5 is connected to two 4s and three 6s)
//\\
4 4
\ /
3

2

1
The sizes of the levels grows in the pattern: three of the same size, then one level with one more, then three of the same size one step larger, etc.
It seems that this idea can be generalized to make many more examples where the degrees increase in levels.