$A$ and $P$ are two 3x3 matrices representing 2d transforms in homogenous coordinates. $A$ is a scaled translation, that is, of the form:
$$ \left[\begin{matrix} A_{00} & 0 & A_{02} \\ 0 & A_{11} & A_{12} \\ 0 & 0 & 1 \end{matrix}\right] $$
$P$ is a general 3x3 matrix which is known to have an inverse $P^{-1}$
Under which circumstances will $PAP^{-1}$ also be a scaled translation?
One approach would be assuming nine variables as the entries of $P$. When working with homogeneous coordinates, you can take $\operatorname{adj}(P)$, the classical adjoint of $P$, instead of the inverse $P^{-1}$ since these only differ by a constant factor. So you compute
$$ B = P\cdot A\cdot\operatorname{adj}(P) $$
The entries of $B$ will have degree $1$ in $A$, but degree $3$ in $P$ (since the adjoint has degree $2$ already). So this would lead to four conditions $… = 0$ and one condition $… \neq 0$ for the bottom right entry.
$$ B_{01} = (P_{00}P_{02}P_{21} - P_{00}P_{01}P_{22})A_{00} + (-P_{01}P_{02}P_{20} + P_{00}P_{01}P_{22})A_{11} + (P_{00}P_{01}P_{20} - P_{00}^2P_{21})A_{02} + (P_{01}^2P_{20} - P_{00}P_{01}P_{21})A_{12} + (P_{01}P_{02}P_{20} - P_{00}P_{02}P_{21}) = 0 \\ B_{10} = (-P_{10}P_{12}P_{21} + P_{10}P_{11}P_{22})A_{00} + (P_{11}P_{12}P_{20} - P_{10}P_{11}P_{22})A_{11} + (-P_{10}P_{11}P_{20} + P_{10}^2P_{21})A_{02} + (-P_{11}^2P_{20} + P_{10}P_{11}P_{21})A_{12} - (P_{11}P_{12}P_{20} + P_{10}P_{12}P_{21}) = 0 \\ B_{20} = (-P_{12}P_{20}P_{21} + P_{11}P_{20}P_{22})A_{00} + (P_{12}P_{20}P_{21} - P_{10}P_{21}P_{22})A_{11} + (-P_{11}P_{20}^2 + P_{10}P_{20}P_{21})A_{02} + (-P_{11}P_{20}P_{21} + P_{10}P_{21}^2)A_{12} - (P_{11}P_{20}P_{22} + P_{10}P_{21}P_{22}) = 0 \\ B_{21} = (P_{02}P_{20}P_{21} - P_{01}P_{20}P_{22})A_{00} + (-P_{02}P_{20}P_{21} + P_{00}P_{21}P_{22})A_{11} + (P_{01}P_{20}^2 - P_{00}P_{20}P_{21})A_{02} + (P_{01}P_{20}P_{21} - P_{00}P_{21}^2)A_{12} + (P_{01}P_{20}P_{22} - P_{00}P_{21}P_{22}) = 0 \\ B_{22} = (-P_{02}P_{11}P_{20} + P_{01}P_{12}P_{20})A_{00} + (P_{02}P_{10}P_{21} - P_{00}P_{12}P_{21})A_{11} + (-P_{01}P_{10}P_{20} + P_{00}P_{11}P_{20})A_{02} + (-P_{01}P_{10}P_{21} + P_{00}P_{11}P_{21})A_{12} - (P_{01}P_{10}P_{22} + P_{00}P_{11}P_{22}) \neq 0 $$
Now if you want to find those $P$ which map any $A$ back to a scaled translation, you'd want to have all of the parentheses equal to zero, except for the constant term in the last inequality which has to be non-zero.
One possible solution would be $P_{01}=P_{10}=P_{20}=P_{21}=0$ with $P_{00}P_{11}P_{22}\neq 0$. That's conjugating your scaled translation with another scaled translation. No surprise there. But that's not the only possible option. One other example would be the rotation by $90°$.
I'd conjecture that any transformation $P$ which fixes the set of all scaled translations under conjugation is either a scaled translation itself or a combination of a scaled translation and a rotation by $90°$. I think this sounds very plausible, but I'm not sure how to prove it.