Let $γ : I → \mathbb{R}^3$ a curve parameterized by arc length. Suppose that the curvature $κ$ never vanishes, so that the Frenet tryro $\{e_1, e_2, e_3\}$ is well defined. Consider a curve $\tilde{γ}: I → \mathbb{R}^3$
defined by $\tilde{γ}(s) = A(γ(s)) + v$, where $A:\mathbb{R}^3 \to \mathbb{R}^3$ is an orthogonal linear application and $v \in \mathbb{R}^3$ is a fixed vector. Denote by $\{ \tilde{e_1}, \tilde{e_2}, \tilde{e_3} \}$, $\tilde{k}$, $\tilde{τ}$ the objects corresponding to $\tilde{γ}$. Show that $\tilde{γ}$ is parameterized by arc length.
$||\tilde{γ}'(s)|| =||A'(γ(s))\cdot γ'(s)||=||A(γ(s))\cdot γ'(s)||$
I'm having a hard time moving on. If it was $||A(γ'(s)))||$ then I would have $||A(γ'(s))||=||γ'(s)|| = 1$ using that $A$ is orthogonal.
You're right, and your only issue seems to be with notation. If $A \colon \mathbb{R}^3 \to \mathbb{R}^3$ is a linear map, then at any point $x \in \mathbb{R}^3$ we have $DA(x) = A$; in other words, the linear map $DA(x)$ happens to be the map $A$ itself. I'm using the notation $DA$, as $A'$ is usually reserved for single variable derivatives.
What you wrote is correct up one minor thing: $$ ||\tilde{γ}'(s)|| = ||DA(γ(s)) \cdot γ'(s)|| = ||A \cdot γ'(s)|| = ||γ'(s)|| = 1. $$ It might be confusing that the application of $A$ is first denoted by $A(x)$ and then by $A \cdot x$, but once you get over this confusion, everything should be fine.