Automorphism groups of odd order This is inspired by this question. Is there a description of finite groups without automorphisms of order $2$?
 A: New version (existence hinted in previous version): If $G$ is a non-trivial finite (solvable) group of odd order with $\Phi(G) = 1$, then $G$ has an automorphism of order $2$.
It is well-known and easy to check that $F = F(G)$ is a product of minimal normal subgroups of $G$, each an elementary Abelian $p_{i}$-group for some prime $p_{i}$.
Also, $F$ is well-known to be complemented in $G$ in this case (I give a proof for completeness:
Choose a proper subgroup $H$ of $G$ minimal subject to $G = FH$ (such exists because $1 \neq F \not \leq \Phi(G)$). Then $(H \cap F) \leq \Phi(H)$ by minimality of $H$. Also $F \cap H$ is normal in $\langle H,F \rangle = G$, since $F$ is Abelian and $F \lhd G$. If $F \cap H \neq 1$, then there is a maximal subgroup $M$ of $G$ with $G = (F \cap H)M$ since $\Phi(G) = 1$. Then $H = (F \cap H)(M \cap H)$ by Dedekind's modular law. But then $H = H \cap M \leq M$ since $F \cap H \leq \Phi(H)$. But then $G = (F \cap H)M \leq M$, contrary to the fact that $M$ is maximal). 
Now $G = FH$ for some subgroup $H$ of ${\rm Aut}(F)$, and the product is semidirect. Thus $G$ is isomorphic to a subgroup of the holomorph $ X = F{\rm Aut}(F)$ (the semidirect product of $F$ with its automorphism group). Here, we have $G \cong  F{\rm Aut}_{G}(F)$, where ${\rm Aut}_{G}(F)$ is the subgroup of ${\rm Aut}(F)$ induced by the conjugation action of $G$ on $F$.
Now let $t$ be the central element of ${\rm Aut}(F)$ of order $2$ which inverts $F$ elementwise (note that $t$ is indeed central in ${\rm Aut}(F)$, because $\alpha(f)^{-1} = \alpha(f^{-1})$ for every $\alpha \in {\rm Aut}(F)$). Then $F\langle t \rangle$ normalizes every subgroup of $X$ containing $F$, so normalizes $F{\rm Aut}_{G}(F) \cong G$.
Now $|(F{\rm Aut}_{G}(F))(F \langle t\rangle)| = 2|F{\rm Aut}_{G}(F)|$, so that $t$ induces an automorphism of order $2$ of $F{\rm Aut}_{G}(F) \cong G$ (recall that $t$ already inverts $F$ elementwise). Note that $F{\rm Aut}_{G}(F)$ is of index $2$ in $(F{\rm Aut}_{G}(F))(F \langle t\rangle)$, so is normal in the latter group.
