I am looking for examples of unbounded operators $T$ on a separable Hilbert space $H$ that satisfies $$\mathrm{Dom}(T)=...=\mathrm{Dom}(T^n)\neq H$$ for given $n>1$, but such that $T^n=\mathrm{id}_{\mathrm{Dom}(T)}$. I cannot find such operators (even for $n=2$) and I am worry if it is possible that they exist (but I don't see the argument why not).
Is it possible to find such $T$ that is also self-adjoint or essentially self-adjoint?
This is a partial answer for the case $n=2$.
Consider the case as stated for $n=2$. And suppose that $T$ is closed, so that its graph $\mathcal{G}(T) = \{ (x,Tx)\in H\times H : x\in\mathcal{D}(T) \}$ is a Hilbert space under the graph inner product $$ \langle x,y\rangle_T = \langle x,y\rangle_H + \langle Tx,Ty\rangle_H. $$ Define $T' : \mathcal{G}(T)\rightarrow \mathcal{G}(T)$ by $T'(x,Tx)=(Tx,T^2x)$, which makes sense because $\mathcal{D}(T)=\mathcal{D}(T^2)$ holds by assumption. It is easy to verify that $T'$ is closed because $T$ is closed. So, by the closed graph theorem, $T'$ is bounded, i.e., \begin{align} \|Tx\|^2+\|T^2x\|^2 & \le \|T'\|^2(\|x\|^2+\|Tx\|^2),\\ \|T^2x\|^2 & \le \|T'\|^2\|x\|^2+(\|T'\|^2-1)\|Tx\|^2. \end{align} So $T : \mathcal{D}(T)=\mathcal{D}(T^2)\subseteq H \rightarrow H$ is bounded. The operator $T$ has a unique continuous extension to the closure $\mathcal{D}(T^2)^c=\mathcal{D}(T)^c$, where the closure is in $H$. This rules out having an unbounded $T$ under the current assumptions. So, in order to get what you want, $T$ cannot be closable. That makes this problem messy.