Does every $p$-group of odd order admit fixed point free automorphisms?
equivalently,
Given an odd order $p$-group $P$, is there a group $C$ such that we can form a Frobenius group $P\rtimes C$?
Note that this is not true for $p$-groups of even order, for example $Q_8$ and $C_4$. But I cannot think of an example with odd order.
The following family of $p$-groups provides counter examples $$G = \langle a, b, c,d |a^{p^n}=b^{p^4}=c^{p^4}=d^{p^2}=1, [a,b]=[a,c]=b^{p^2}, [a,d]=c^{p^2}, [b,c]=a^{p^{n-2}}, [b,d]=c^{p^2}, [c,d]=c^{p^2} \rangle$$ with $p$ odd, and $n>3$.
For such a group $G$, every automorphism is central, that is $\operatorname{Aut}(G)$ acts trivially on $G/\operatorname{Z}(G)$. It is easy to see that the number of central automorphisms in that case (actually, for any finite group with no abelian direct factor) is equal to the order of $\operatorname{Hom}\left(G/G',\operatorname{Z}(G)\right)$. Thus $\operatorname{Aut}(G)$ is a $p$-group, so there is no automorphism acting fixed point freely on $G$.
The above example is due to V. Jain, P. Rai and M. Yadav.
As mentioned by Steve D, it is proved by U. Martin and G. Helleloid that (in some sense) almost finite $p$-groups have an automorphism group of $p$-power order, thus 'almost' $p$-groups have no fixed point free automorphisms.