Show that exists $A\in SO(3)$ and $b\in \mathbb{R}^3$ such that $\tilde{\gamma} = A\gamma + b$ I want to show that 
if $\tilde{\gamma} : I \rightarrow \mathbb{R}^3$ and $\gamma : I \rightarrow \mathbb{R}^3$ have same curvature $\kappa$ and torsion $\tau$,
there exists $A\in SO(3)$ and $b\in \mathbb{R}^3$ such that $\tilde{\gamma} = A\gamma + b$.
I'm really confused about this.
Please note that it is not to prove 
if $\tilde{\gamma} = A\gamma + b$,  then they have same curvature $\kappa$ and torsion $\tau$.
 A: Assume each curve is parametrized according to arc length $s$.  
For each $s$, the Frenet-Serret frame $\mathbf{T}(s)$, $\mathbf{N}(s)$, $\mathbf{B}(s)$ (unit tangent, normal, and binormal vectors) are an orthonormal set.  This means that the matrix $F(s) = \begin{bmatrix} \mathbf{T}(s) & \mathbf{N}(s) & \mathbf{B}(s) \end{bmatrix}$ is in $SO(3)$ for each $s$.  Letting $\tilde{\mathbf{T}}(s)$, $\tilde{\mathbf{N}}(s)$, and $\tilde{\mathbf{B}}(s)$ be the analogous frame for $\tilde\gamma$, and $\tilde F(s)$ the associated matrix, we define $A(s) = \tilde F(s) F(s)^{-1}$.  This is also in $SO(3)$ for each $s$.
The inverse of a matrix in $SO(3)$ is equal to its transpose.  So we can compute $A(s)$ explicitly in terms of the vectors:
\begin{align*}
    A(s) &= \begin{bmatrix} \tilde{\mathbf{T}}(s) & \tilde{\mathbf{N}}(s) & \tilde{\mathbf{B}}(s) \end{bmatrix}\begin{bmatrix} \mathbf{T}(s) & \mathbf{N}(s) & \mathbf{B}(s) \end{bmatrix}^{-1} \\
         &= \begin{bmatrix} \tilde{\mathbf{T}}(s) & \tilde{\mathbf{N}}(s) & \tilde{\mathbf{B}}(s) \end{bmatrix}\begin{bmatrix} \mathbf{T}(s) & \mathbf{N}(s) & \mathbf{B}(s) \end{bmatrix}^{T} \\
         &= \begin{bmatrix} \tilde{\mathbf{T}}(s) & \tilde{\mathbf{N}}(s) & \tilde{\mathbf{B}}(s) \end{bmatrix}\begin{bmatrix} \mathbf{T}(s)^T \\ \mathbf{N}(s)^T \\ \mathbf{B}(s)^T \end{bmatrix} \\
         &= \tilde{\mathbf{T}}(s)\mathbf{T}(s)^T + \tilde{\mathbf{N}}(s)\mathbf{N}(s)^T + \tilde{\mathbf{B}}(s)\mathbf{B}(s)^T
\end{align*} 
Here we are taking the outer product of each tilde vector with the other non-tilde vector, rather the dot (or inner) product.  So $\tilde{\mathbf{T}}(s)\mathbf{T}(s)^T$ is a rank-one $3\times 3$ matrix for each $s$, to which we add two more rank one matrices.
We claim that $A(s)$ is a constant function.  This comes from the Frenet-Serret formulas:
\begin{align*}
    \mathbf{T}'(s) &= \kappa \mathbf{N}(s) \\
    \mathbf{N}'(s) &= -\kappa\mathbf{T}(s) +\tau \mathbf{B}(s) \\
    \mathbf{B}'(s) &= -\tau \mathbf{N}(s)
\end{align*}
By assumption, the same is true for the tilde frame.  So by the product rule:
\begin{align*}
    A' &= \tilde{\mathbf{T}}'\mathbf{T}^T + \tilde{\mathbf{T}}(\mathbf{T}')^T
         +\tilde{\mathbf{N}}'\mathbf{N}^T + \tilde{\mathbf{N}}(\mathbf{N}')^T
         +\tilde{\mathbf{B}}'\mathbf{B}^T + \tilde{\mathbf{B}}(\mathbf{B}')^T \\
       &= (\kappa \tilde{\mathbf{N}})\mathbf{T}^T 
         + \tilde{\mathbf{T}}(\kappa\mathbf{N})^T
         +(-\kappa\tilde{\mathbf{T}} + \tau \tilde{\mathbf{B}})\mathbf{N}^T
         +\tilde{\mathbf{N}}(-\kappa{\mathbf{T}} + \tau {\mathbf{B}})^T
         +(-\tau\tilde{\mathbf{N}})\mathbf{B}^T
         +\tilde{\mathbf{B}}(-\tau{\mathbf{N}})^T
\end{align*}
Distribute the transpose and collect like terms; you get zero.
We're not quite there yet, but that is your matrix $A$.  I'll leave it to you to figure out how to find the $b$.
