Q. I have a finite group $G$, and a subset $H$ given by $$H=\{h\in G~:~O(g)~\text{is odd}\},$$
My doubt: Can we say that $H$ form a normal subgroup of $G$ in the light of order preserving property of conjugate isomorphism?
But I have to ensure $H$ is a subgroup, If not, what about the normality of the subgroup generated by $H$?
You're being asked about the subgroup generated by the elements of odd order (the set you propose is a not a subgroup in general, try with $S_4$).
Since the inverse of an element of odd order has odd order, the subgroup you want consists of the elements of the form $$ a_1a_2\dots a_n $$ where each $a_i$ has odd order.
A conjugate of this element would be $$ g(a_1a_2\dots a_n)g^{-1}=(ga_1g^{-1})(ga_2g^{-1})\dots(ga_ng^{-1}) $$ and each $ga_ig^{-1}$ has odd order because conjugation by $g$ is an automorphism of the group.