$M_2= \langle(\sqrt{2}^{\sqrt{2}},\infty),\cdot \rangle$ , $M_1 = \langle(2,\infty),\cdot\rangle$
attempt:
for all sign of function $f$
1.$h(F^{M}(a_1,a_2,...a_n))=F^N(h(a_1),h(a_2),...(h(a_n))$
2.Injective function
3.Surjective function
if i try to prove isomorphic in B. between $M_2$ , $M_1$ i dont know how to prove them with in open segment
Let $\alpha,\beta\in\mathbb{R}$ such that $\alpha,\beta>1$. Then we claim $M_\alpha:=\langle (\alpha,\infty),\cdot\rangle$ and $M_\beta:=\langle (\beta,\infty),\cdot\rangle$ are isomorphic. First note that $M_\alpha\cong\langle (\ln\alpha,\infty),+\rangle$ and $M_\beta\cong\langle (\ln\beta,\infty),+\rangle$, both via the map $x\mapsto\ln x$. Now we define an isomorphism $$f:\langle (\ln\beta,\infty),+\rangle\rightarrow \langle (\ln\alpha,\infty),+\rangle$$ via $x\mapsto\frac{\ln \alpha}{\ln \beta}x$. Since $\ln\alpha,\ln\beta>0$, this is a well-defined bijection, and, letting $\lambda=\frac{\ln{\alpha}}{\ln\beta}$, we have $f(x+y)=\lambda(x+y)=\lambda x+\lambda y=f(x)+f(y)$, so that $f$ is a $\{+\}$-embedding and hence an isomorphism, as desired.
Now, applying this to the case $\alpha=2$ and $\beta=\sqrt{2}^\sqrt{2}$ gives the desired result.
The map $f : x \mapsto x^{\sqrt 2}$ is an homomorphism from $M_2= \langle(\sqrt{2}^{\sqrt{2}},\infty),\cdot \rangle$ onto $M_1 = \langle(2,\infty),\cdot\rangle$.
as for any $x,y \gt 0$ we have $(x \cdot y)^{\sqrt{2}} = x^{\sqrt{2}} \cdot y^{\sqrt{2}}$. Also
$$f\left( \sqrt 2^{\sqrt 2}\right)=\left(\sqrt 2^{\sqrt 2}\right)^{\sqrt 2} = (\sqrt 2)^2 = 2$$
and as $f$ is continuous strictly increasing it is bijective. Finally $f$ is an isomorphism between the two structures.
