I was asked to show, given an isometry U from a finite dimensional inner product space V to itself that $$ A_n(x)=\frac{1}{n}\sum_{0}^{n}U^{n}(x) \to 0 \text{ as } n\to \infty \text{ for} x\in Im(I-U) $$ so letting $ y-U(y)$ $$ A_n(x)=\frac{1}{n}\sum_{0}^{n}U^{n}(y-U(y))=\frac{1}{n}\sum_{0}^{n}U^{n}(y)-U^{n+1}(y)=\frac{U(y)-U^{n+1}(y)}{n} $$ and so clearly for a central isometry $$ |A_n|\leq \frac{|U(y|+|U^{n+1}(y)|}{n}=\frac{2|U(y)|}{n}\to 0 \text{ as } n \to \infty $$ as required
For a general isometry $U(x)=Ax+v$ things are less straightforward, in fact it seems to me that taking $U:\mathbb{R}^{n}\to\mathbb{R}^{n}$ s.t $U(x)=x+v$, say, then $$ A_n(-v)=\frac{1}{n}\sum_{0}^{n}U^{n}(0-U(0))=\frac{U(0)-U^{n+1}(0)}{n}=\frac{-v-nv}{n}\to -1 \text{ as } n \to \infty $$ Have I messed up somewhere with my counter-example or is the result indeed only true for central isometries?
You are correct. I would add that usually when people refer to "isometries" of inner product spaces (or more generally, of normed vector spaces), they mean only linear maps which are isometries, not arbitrary maps. To refer to possibly non-linear isometries, you usually say something like "affine isometries".