Let there be two sets $A, B\subseteq\Bbb{R}$ and let there be two binary operations $*_M$ and $*_N$. Under what circumstances is $(A,*_M)\cong(B,*_N)$?
I have found a couple of general working cases. If $a*_Nb=(a^n*_Mb^n)^{1/n}$, then $(\Bbb{R}^+,*_M)\cong(\Bbb{R}^+,*_N)$ using the function $f(a)=a^n$. Also, using similar logic, the function $f(x)=\ln(x)$ provides $(\Bbb{R},*_M)\cong(\Bbb{R}^+,*_N)$, when $a*_Nb=e^{\ln(a)*_M\ln(b)}$.
Do isomorphisms only occur if, given proper $A,B$, when $a*_Mb=f^{-1}\Big(f(a)*_Nf(b)\Big)$?
Given that when trying to find an isomorphism between(both are on $\Bbb{R}$): $a*_Mb=a+b+ab$ and $a*_Mb=a^2+b^2$, you get the equation $f(a)^2+f(b)^2=f(a+b+ab)$, which has solutions $f(x)=0$ and $f(x)=\pm{1\over2}$. While I know $f(x)=0$ is a trivial solution for any two magma, $f(x)={1\over2}$ is not, and does reflect something about the nature of the equation. I know that $f(x)$ must be a bijection for $(A,*_M)\cong(B,*_N)$, so is there anything the existence of this function shows other than just both $*_M$ and $*_N$ are quadratic?