Take the natural graph theoretic interpretation, and let $A$ be one of the vertices. Also, let $B = B_1,\cdots,B_k$ be the set of vertices adjacent to $A$. First, note that the subgraph consisting of the vertices in $B$ consists of $\frac{k}{2}$ disjoint edges (note that this implies $B$ is even), as if we pick $X \in B$, then there must be a vertex $Y \in B$ such that $XY$ is a drawn edge, and there can be no other edge in $B$ connected to $X$ or $Y$, as that would contradict uniqueness. Then we can choose another vertex in $B$ and repeat this procedure to imply the result. Now, note that every other vertex $T$ in the graph (not in $A$ or $B$) must be connected to exactly one vertex in $B$, by considering $AT$. We can find sets of adjacent vertices $S_1,S_k$ of vertices such that a vertex $X$ is in $S_i$ if and only if $X$ is connected to $B_i$. WLOG $B_1$ is connected to $B_2$, $B_3$ is connected to $B_4$ and so on. Then pick a vertex $Q$ in $B_1$. Note that for each $i>2$, $Q$ must be connected to a vertex in $B_i$, by considering edge $QB_i$ and noting that the only vertices connected to $B_i$ are in $S_i$. However, this means $Q$ has at least degree $k$, as it is also connected to $B_1$ and another vertex in $S_1$, and there are $k-2$ $S_{i>2}$s. Therefore, given any vertices $A$,$B$ not connected to each other, $deg(A) \ge deg(B)$ and $deg(B) \ge deg(A)$, implying that $deg(A) = deg(b)$, as required.

It can be shown from here without too much effort that either $deg(A) = deg(B)$ for all $A,B$ in the graph, or there is one vertex connected to every other vertex. Also, it is relatively easy to show that if $x_1,\cdots,x_n$ are the degrees of the graph then $\sum \frac{x_i(x_1-1)}{2} = \frac{n(n-1)}{2}$, which would imply that if all of the degrees are equal then $n-1$ can be expressed as $x(x-1)$ for some $x$ or there is a vertex connected to every other vertex. I suspect that the last case is impossible, but cannot prove it. Maybe someone smarter than me can help?