%
\label{thm:rigidity}
The automorphism group of $\tau$ is trivial:
\[
    \boxed{\Aut(\tau) = \{\id\}.}
\]

Let $\varphi : \Obj(\tau) \to \Obj(\tau)$ be an automorphism.

\medskip
\textbf{Step 1: $\varphi$ fixes $\omega$.}

Since $\omega$ is a named constant in the signature,
any $\Sigma_\tau$-automorphism must fix it:
$\varphi(\omega) = \omega$.

\medskip
\textbf{Step 2: $\varphi$ fixes each generator.}

Similarly, $\varphi(\alpha) = \alpha$,
$\varphi(\pi) = \pi$,
$\varphi(\gamma) = \gamma$,
$\varphi(\eta) = \eta$.
All five generators are fixed by $\varphi$.

\medskip
\textbf{Step 3: $\varphi$ fixes all orbit elements.}

Let $x \in O_g$ for some $g \in \{\alpha, \pi, \gamma, \eta\}$.
Then $x = \rho^n(g)$ for a unique $n \geq 0$
(by the Ontic Closure Theorem, Part~(4)).
Since $\varphi$ commutes with $\rho$ and fixes~$g$:
\[
    \varphi(x) = \varphi(\rho^n(g))
    = \rho^n(\varphi(g)) = \rho^n(g) = x.
\]
The second equality uses induction on $n$:
$\varphi(\rho^0(g)) = \varphi(g) = g = \rho^0(g)$,
and if $\varphi(\rho^k(g)) = \rho^k(g)$ then
$\varphi(\rho^{k+1}(g)) = \varphi(\rho(\rho^k(g)))
= \rho(\varphi(\rho^k(g))) = \rho(\rho^k(g)) = \rho^{k+1}(g)$.

\medskip
\textbf{Step 4: Conclusion.}

Every element of $\Obj(\tau)$ is either $\omega$
(fixed by Step~1) or an orbit element $\rho^n(g)$
(fixed by Steps~2--3).
Therefore $\varphi = \id$.