Let $H$ be a $\mathbb C$-Hilbert space, $T$ be a densely-defined linear operator on $H$ and $\Gamma(T)$ denote the graph of $T$. By definition, $T$ is symmetric if $T\subseteq T^\ast$ and self-adjoint if $T=T^\ast$. Let $$U:H\oplus H\to H\oplus H\;,\;\;\;(x,y)\mapsto(y,-x).$$ We know that $T^\ast$ is closed and $$\Gamma(T^\ast)=(U\Gamma(T))^\perp.\tag1$$ We immediately conclude that $T$ is self-adjoint iff $$\Gamma(T)^\perp=U\Gamma(T)\tag2$$ (since in that case $\Gamma(T)=\Gamma(T^\ast)$ is closed and hence $\overline{U\Gamma(T)}=U\overline{\Gamma(T)}=U\Gamma(T)$).
Question 1: Can we give an analogous equivalent condition for $T$ being symmetric?
Question 2: Can we show that $T$ is symmetric if and only if $T$ is closable and $\overline T=T^{\ast\ast}\subseteq T^\ast$?