Show for isometry $T:V \rightarrow V$, $\frac{1}{n}\sum_{i=0}^{n-1} T^i$ converges to the orthogonal projection on $ker(T-I)$.
For each $x \in V$, write $x = y + z$ where $y \in \ker(T-I)$ and $z \perp \ker(T-I)$. Then
Since $Ty = y$, we get $Q_n(y) = y$.
Note that $z \in \ker(T-I)^{\perp} = \operatorname{im}(T-I)^*=\operatorname{im}(T^{-1}-I)$. In the last step, we utilized the fact that $T$ is invertible and $T^* = T^{-1}$. So there exists $w \in V$ such that $z = (T^{-1}-I)w$. Then
$$ \| Q_n(z) \| = \left\| \frac{1}{n}(T^{-1}w - T^{n-1}w) \right\| \leq \frac{\|T^{-1}w\| + \|w\|}{n} \xrightarrow[n\to\infty]{} 0 $$
and so, $Q_n(z) \to 0$.
Combining altogether, it follows that $Q_n(x) \to y$ as $n\to\infty$. Since $y$ is the orthogonal projection of $x$ onto $\ker(T-I)$, we are done.