This was a comment to the answer [here](https://mathoverflow.net/questions/351715/automorphism-groups-of-odd-order) . It is one of the series of questions about finite groups with automorphism groups of odd order and would reduce the question to nilpotent groups. 


**Question.** Is it true that a finite group $G$ has an automorphism of order 2 iff the Fitting subgroup $F(G)$ has an automorphism of order $2$.


Geoff Robinson proved in a comment [here](https://mathoverflow.net/questions/351715/automorphism-groups-of-odd-order) that one of the implications is true if $G$ has odd order. He doubts that "iff" is true but no example was given. 



Automorphism groups of a group and of its Fitting subgroup
It can happen that the whole group has no automorphism of order 2 although the Fitting subgroup does:



> 
> Let 
> 
> 
> * $N$ be a non-abelian 2-generated $p$-group for some prime $p$, such that $N/Z(N)$ has no automorphism of order $2$;
> * $L$ a nontrivial group of order coprime to $2p$ with no automorphism of order $2$.
> 
> 
> I claim that the wreath product $$G=N\wr L= N^L\rtimes L$$
>  has no automorphism of order $2$, although its Fitting subgroup $N^L$ obviously does (permute 2 factors).
> 
> 
> 


Indeed, let $s$ be an automorphism with $s^2=1$. Then $s$ preserves the Fitting subgroup. Also it preserves some conjugate of $L$: indeed, consider a $2|L|$-Hall subgroup of $G\rtimes\langle s\rangle$, and let $L\_1$ be its subgroup of index $2$: then $L\_1$ is a $|L|$-Hall subgroup of $G$ and hence is conjugate to $L$. So henceforth assume that $s$ preserves $L$.


Since $L$ has no automorphism of order $2$, $s$ acts as the identity on $L$. Hence on $N^L$, $s$ commutes with the $L$-action.


Let now $t$ be an arbitrary automorphism of $N$ commuting with the $L$-action. Let $N\_u$ be the copy of $N$ indexed by $u\in L$ in $N^L$. For $u\in L$ let $M\_u$ be the projection of $t(N\_1)$ in $N$ (through the $u$-th projection to $N\_u$). Then for $v\in L$, $t(N\_v)=t(vN\_1)=v(tN\_1)$, so the $u$-projection of $t(N\_v)$ is equal to $M\_{v^{-1}u}$. Since $N\_1$ and $N\_v$ commute for $v\neq 1$, we deduce that $M\_u$ and $M\_{v^{-1}u}$ commute for $v\neq 1$.


Also since the $N\_u$ generate $N^L$, the $M\_u$ generate $N$. Now we will use that $N$ is 2-generated (so $N$ is a $p$-group for some odd prime $p$ and $N/[N,N]N^p$ has order $p^2$). If there exists $u$ such that $M\_u=N$, we deduce that $M\_v$ is central in $N$ for all $v\neq u$. Otherwise, there exist $u\neq v$, $x\in M\_u$ and $y\in M\_v$ such that $\{x,y\}$ generates $N$; since $x,y$ commute, it follows that $N$ is abelian, contradiction.


Hence, modulo the center, $\mathrm{Aut}(N^L)$ permutes the factors (I used the 2-generation condition there). Now if we restrict to automorphisms commuting with the $L$-action, this action on $L$ is through the centralizer of the action of $L\_{\mathrm{left}}$, which is $L\_{\mathrm{right}}$.


Applied to $L$, this shows that for every $u\in L$, $s(N\_u)$ equals $N\_u$ modulo $Z(N^L)$. Composed with the projection to $N\_u$, this yields an endomorphism $s\_u$ of $N\_u$, and we have $s\_u^2$ induces trivial automorphism of $N\_u/Z(N\_u)\simeq N/Z(N)$. Hence $s\_u$ induces an automorphism of order $\le 2$ of $N/Z(N)$, and by assumption, this implies that $s\_u$ is trivial modulo the center. Hence $s$ is trivial modulo the center. But this implies that $s$ has order $p^k$ dividing the exponent of $Z(N)$. So $s$ is trivial.


