Let $f:V\to V$ be an isometry on a finite-dimensional vector space $V$. Then there exists a unique $u\in V$, and a unique orthogonal operator $A:V\to V$ such that $f = T_u\circ A$. This helps conclude that all isometries on $V$ can be written in this form.
I will denote the translation map which sends all $v\in V$ to $v+u\in V$ by $T_u$, i.e. $T_u(v) = u+v$.
Using this post as a Lemma, we see that the map $f - f(0)$ is orthogonal. Let's say $u = f(0) \in V$. So, $f-f(0) = T_{-u}\circ f = A$ where $A$ is an orthogonal map on $V$. So we have $f = T_u\circ A$.
How do I prove the uniqueness of $u\in V$ and $A\in\mathcal{L}(V,V)$?
Usually, we work with cases where we have to show the uniqueness of only one of the variables, however, it's a little confusing here since we have to show the uniqueness of $u$ and $A$ both. To begin, I would assume that the isometry $f$ on $V$ can be written as $f = T_{u_1}\circ A_1 = T_{u_2}\circ A_2$ for some $u_1,u_2\in V$ and orthogonal $A_1,A_2\in\mathcal{L}(V,V)$. What next?
Thanks!
If $f = T_{u_1} \circ A_1 = T_{u_2} \circ A_2$, then we have $$ f(0) = T_{u_1}(A_1(0)) = T_{u_1}(0) = u_1, $$ and similarly $f(0) = T_{u_2}(0) = u_2$. Thus, $u_1 = u_2$. From there, we have $$ A_1 = T_{u_1}^{-1} \circ f = T_{u_2}^{-1} \circ f = A_2. $$