Are there isomorphic groups between $\Bbb{A}_4\times \Bbb{Z}_3,\Bbb{D}_{18},\Bbb{D}_{9}\times \Bbb{Z}_2,\Bbb{S}_3\times \Bbb{S}_3$? Where $\Bbb{D}$ is the dihedral group,$\Bbb{A}$ alternating group and $\Bbb{S}$ is the symmetric group.
I've tried the things below but I feel like I'm doing things wrongly and I don't know what to do with $\Bbb{A}_4\times \Bbb{Z}_3$.
For $\Bbb{D}_{18}$ and $\Bbb{D}_{9}\times \Bbb{Z}_2$ I've tried constructing $f$ such that $f(\sigma\rho^k) = (\sigma\rho^{(k-k\bmod 2)/2},k\bmod 2)$ and $f(\rho^{2k})=(\rho^k,0)$ and $f(\rho^{9})=(1,1)$ and $f(\rho)=(\rho,1),f(\rho^3)=(\rho^2,1),f(\rho^5)=(\rho^3,1),f(\rho^7)=(\rho^4,1),f(\rho^{11})=(\rho^5,1),f(\rho^{13})=(\rho^6,1),f(\rho^{15})=(\rho^7,1),f(\rho^{17})=(\rho^8,1)$ which should be an isomorphism but I couldn't prove it.
For $\Bbb{D}_{18}$ and $\Bbb{S}_3\times \Bbb{S}_3$, $\Bbb{D}_{18}$ has a element of order $18$ while $\Bbb{S}_3\times \Bbb{S}_3$ has at most an element of order $9$. Similar with $\Bbb{D}_{9}\times \Bbb{Z}_2$.
Your idea for an isomorphism $\Bbb{D}_{18}\cong\Bbb{D}_9\times\Bbb{Z}_2$ is good. What part of proving this is an isomorphism is a problem? Try writing out what $f$ does to an element $\sigma^a\rho^b\in\Bbb{D}_{18}$ in general, in stead of defining it for each element separately. That will surely make it more clear why this should be an isomorphism.
Your last consideration is a good one; the groups $\Bbb{D}_{18}$ and $\Bbb{S}_3\times\Bbb{S}_3$ are indeed not isomorphic because $\Bbb{D}_{18}$ has an element of order $18$ while $\Bbb{S}_3\times\Bbb{S}_3$ does not. Similar arguments show that no two of these groups are isomorphic, except $\Bbb{D}_{18}$ and $\Bbb{D}_9\times\Bbb{Z}_2$. As this seems like a homework exercise, I'll leave the rest for you to figure out (for now).