Let $k$ be a field of characteristic zero (I do not mind to assume that $k=\mathbb{C}$, if this helps).
Lüroth theorem says that a field $L$, $k \subset L \subset k(t)$ containing a nonconstant polynomial over $k$ is equal to $k(h)$ for some $h \in k[t]$. More precisely, by Lüroth theorem there exists such $h \in k(t)$, and E. Noether has proved that actually one can take $h \in k[t]$ (see "Selected Topics on Polynomials" by A. Schinzel).
Let $f=f(t),g=g(t) \in k[t]-k$, and $L:=k(f,g)$. By Lüroth-Noether theorem we have $k(f,g)=k(h)$, for some $h=h(t) \in k[t]$. Further assume that $f$ and $g$ are odd polynomials, namely, $f=a_{2n+1}t^{2n+1}+a_{2n-1}t^{2n-1}+\cdots+a_3t^3+a_1t$ and $g=b_{2m+1}t^{2m+1}+b_{2m-1}t^{2m-1}+\cdots+b_3t^3+b_1t$, $a_i,b_j \in k$.
Is it true that $h$ must be an odd polynomial too? If this is too much to ask, than at least is it true that $h$ must have an odd degree? (though it can have monomials of even degrees).
For example: $f=t^3, g=t^7$. Then $\frac{g}{f^2}=\frac{t^7}{t^6}=t$, and then $L=k(t)$, so $h=t$ is also an odd polynomial. Of course, this is just a simple example which may not hint about a general solution.
Another example: $f=t^3, g=t^7+t^5$. Then $\frac{g}{f}=\frac{t^7+t^5}{t^3}= \frac{t^3(t^4+t^2)}{t^3}=t^4+t^2$. (Notice that $\frac{g}{f^2}=\frac{t^7+t^5}{t^6}=t+\frac{1}{t}$, but I do not know if this helps). What is $h$?
A trivial remark: Since $k(h)=k(h+\lambda)$ for any $\lambda \in k$, it is no loss of generality to assume that $h(0)=0$ , so we avoid trivial counterexamples for which $h(0)\neq 0$ (such $h$ is not an odd polynomial).
Any hints and comments are welcome!