Hint:
Given a self complementary graph on $n$ vertices, try to construct a self-complementary graph on $n+4$ vertices.
Elaboration on the hint:
Given a self-complementary graph $G$ with $n$ vertices, add 4 more vertices which already form a 'Z' graph (the self-complementary graph for $n=4$). Connect the vertices of $G$ to some of four newly added vertices. Show that the new graph so formed, is a self-complementary graph on $n+4$ vertices.
Full answer:
Connect every vertex of $G$ to two of the vertices which are the farthest apart on the the n=4 graph (which is basically a path, and you connect all vertices of $G$ to the end points of that path).
This works even when $G$ has only one vertex.
Let $H$ and $K$ be the partite sets of the bipartition of $G-u$. Let $w$ be a vertex different from both $u$ and $v$; without loss of generality we may assume that $w\in H$. Now remember that $v$ was a cut vertex of $G$, so that $G-v$ has at least two components. Let $C$ be the component of $G-v$ that contains $w$.
Suppose that $x$ is a vertex of $G-v$ that is not in $C$. Then there is no edge between $w$ and $x$ in $G-v$, since there are no edges between different components of $G-v$. This means that there is no edge between $w$ and $x$ in $G$, since removing $v$ from $G$ would not have affected such an edge. $H$ and $K$ are the partite sets of a spanning biclique for $G-u$, so if $x$ were in $K$, $G-u$ would have an edge between $w$ and $x$, and hence so would $G$. We just saw that this is not the case, so $x$ cannot be in $K$ and must therefore be in $H$. This shows that every vertex of $G-v$ that is not in $w$’s component $C$ of $G-v$ must belong to the partite set $H$, the one that contains $w$.
Now what about a vertex $x$ of $G-v$ that is in $C$ (but isn’t $w$)? Let $y$ be any vertex of $G-v$ that is not in $C$. There is no edge in $G-v$ between $x$ and $y$, since they are in different components, and since neither of them is $v$, there is no edge between them in $G$, either. But then there is no edge between them in $G-u$, which means that they must be in the same partite set of the spanning biclique. We just showed that $y$ is in $H$, so $x$ is also in $H$.
We have now shown that every vertex of $G-u$ that is also in $G-v$ must belong to $H$, the same partite set that contains $w$. On the other hand, $K\ne\varnothing$. The only vertex of $G-u$ that isn’t in $G-v$ is $v$ itself, so $v$ must be the one vertex in $K$: $K=\{v\}$, which implies that $v$ is adjacent to every vertex in $H$. $H$ contains every vertex of $G-u$ except $v$, so $|H|=n-2$, and therefore $\deg_Gv\ge n-2$. But then
$$\deg_{G'}v=(n-1)-\deg_Gv\le(n-1)-(n-2)=1\;:$$
$v$ has degree at most $1$ in $G'$. Since $G$ is self-conjugate, $G$ also has some vertex of degree at most $1$.
Finally, if $\deg_{G'}v$ were $0$, $G$ would also have a vertex of degree $0$, so neither $G$ nor $G'$ would be connected. This, however, is impossible: for any graph $G$, at least one of $G$ and $G'$ is connected. Thus, $\deg_{G'}v=1$, and $G$ also has a vertex of degree $1$.
Best Answer
A very nice problem. I did not know it, but I find it hard to give a hint without giving away the full proof. Here is a try: consider the adjacency of the vertex of degree 1 and the vertex of degree $n-2$.
Here is the proof then.
Suppose there is only one vertex $v$ of degree 1. Then there is also only one vertex $w$ of degree $n-2$. Let $\phi$ be an isomorphism mapping $G$ to its complement. Clearly $\phi(v)=w$ and $\phi(w)=v$. But this implies that $v$ and $w$ are adjacent in $G$, if and only if they are adjacent in the complement.
This contradiction shows that there must be another vertex of degree 1.