I want to know if this statement is true.
If $f \in I(\mathbb{R}^2)$ and $T_c$ is a translation, then ${T_c}^{-1} \cdot f \cdot T_c$ is the same kind of isometry as $f$.
This question arises from a proof of the statement "If $f \in I(\mathbb{R}^2)$, then $f$ is either translation, rotation, reflection, or glide."
The proof that I'm reading (I have no link since it is from my lecture note) uses the strategy as follows.
$f(x)=Ax+b$ for some $A \in O(2)$ and $b \in \mathbb{R}^2$.
i) $1$ is an eigenvalue of $A$, then ${T_c}^{-1} \cdot f \cdot T_c=A$ is either reflection or glide. Thus, $f$ is either reflection or glide. ($A=Id$ gives translation).
ii) $1$ is not an eigenvalue of $A$, then ${T_c}^{-1} \cdot f \cdot T_c=A$ is either rotation or reflection. Thus, $f$ is either rotation or reflection.
Here, $c$ (for translation $T_c, {T_c}^{-1}$) depends on each case, but I believe this wouldn't matter for this question.
If we can conclude as above, I think it must be true that "if $f \in I(\mathbb{R}^2)$ and $T_c$ is a translation, then ${T_c}^{-1} \cdot f \cdot T_c$ is the same kind of isometry."
For example, if ${T_c}^{-1} \cdot g \cdot T_c$ is some glide, then $g$ is some glide.
Is my assumption correct?
Edit:
I found an article about the conjugacy classes of Euclidean Group.
I believe my assumption is the same as the idea of the article, correct? However, if I could learn any small details (which the article lacks), I would be really grateful.