Let $H$ be a Hilbert space and $A:D(A)\subset H \rightarrow H$ be symmetric and closed.
Assume $A$ has a selfadjoint extention $B$. Then the Cayley transform of $A$ has also a unitary extention i.e. the Cayley transform of $B$ which is $C_B:=(B-i)(B+i)^{-1} \in L(H)$.
Furthermore we have that $U:=C_B \upharpoonright_{\mathscr{K}_-} : \mathscr{K}_-\rightarrow \mathscr{K}_+ \in L(\mathscr{K}_-,\mathscr{K}_+)$ is unitary, where $\mathscr{K}_\pm:=\mathrm{ran}(A\pm i)^\perp$.
Hence the deficiency indeces $n_\pm(A):=\mathrm{dim}\left(\mathscr{K}_\pm\right)$ are the same, since $U$ is an isometric isomorphism ($\Leftrightarrow$ unitary).
Assume the converse is true. Thus assume given $A:D(A)\subset H \rightarrow H$ (symmetric and closed) $n_+(A)=n_-(A)$.
How can we find a unitary extention $U\in L(H)$ of $C_A:=(A-i)(A+i)^{-1}$?
That would imply that $A$ has a self adjoint extention.
My attempt so far is:
According to: Hilbert spaces have same dimension iff their isomorphic
We can extend $C_A: \mathrm{ran}(A+i)\rightarrow \mathrm{ran}(A-i)$ on the complement space $\mathrm{ran}(A+i)^\perp$ by an other isometry $I:\mathrm{ran}(A+i)^\perp\rightarrow \mathrm{ran}(A-i)^\perp$ and would obtain an isometry on the whole space $U$ be setting $U\varphi=I\varphi$ for $\varphi\in \mathrm{ran}(A+i)^\perp$ and $U\varphi=C_A\varphi$ if $\varphi\in \mathrm{ran}(A+i)$. Assume this isometry is onto then $U$ is unitary and $I-U$ has dense range since $\mathrm{ran}(I-C_A)=D(A)$ is dense in $H$. Thus $U$ is the Cayley transform of $B:=i(I+U)(I-U)^{-1}$ that extends $C_A$. Thus $B$ is a self adjoint extention of $A$.
What I have to settle is if $U$ is indeed onto, because I just assumed it. And I did not checked the proof of the assertion from the link.