A finite 2-group acts on an elementary abelian group of odd order I would be grateful if you have an idea how to prove the following:
Let a finite $2$-group $P$ act on an elementary abelian group of odd order $N$ such that the centralizer $C_P(N)=1$, and then form the semi-direct product $G:=NP$. An element $g\in G$ is said to be real if $g$ is $G$-conjugate to $g^{-1}$. Is it true that
The number of nontrivial real elements of G inside $N$ is at least $|P:P_2|$, where $P_2$ is the subgroup of $P$ generated by all elements $x^2$ with $x\in P$? 
PS: I am able to prove this when $P$ is cylic. I hope that the general statement is correct.
 A: Note that your $P_{2}$ is the Frattini subgroup $\Phi(P)$ since $P/P_{2}$ is an elementary Abelian $2$-group while also $P_{2} \leq \Phi(P)$ by the way you defined it.
Now $N = [N,P] \times C_{N}(P)$ and $P$ acts faithfully on $[N,P]$, so we might as well suppose that $C_{N}(P) =1$, and we do so. Now write $N = N_{1} \times N_{2} \times \ldots N_{r} $ where each $N_{i}$ is minimal normal, and set $K_{i} = C_{P}(N_{i})$. Then each element of $N_{i}$ is inverted by a central involution of $P/K_{i}$ by irreducibility.
If $K_{i} = K_{j}$ for some $i <j$, we can delete $N_{j}$ from the direct product, and still have faithful action of $P$. So we may assume that $K_{i} \not \cong K_{j}$ when $i \neq j,$ and we do so. In fact, we may assume that $\prod_{j \neq i} N_{j}$ is not faithful for any $i$ by discarding redundant factors while retaining faithful action. Let $X_{i} = C_{P} (\prod_{j \neq i} N_{j})$ for each $i.$
Set $Q = X_{1} \times X_{2}\times  \ldots \times X_{r}$, which is a direct product. 
Then $Q \lhd P$ , and for each $i$, we may choose an element $u_{i} \in X_{i}$ which inverts every element of $N_{i}$, (and centralizes $N_{j}$ whenever $j \neq i$).
Then $u_{1}u_{2}\ldots u_{r}$ inverts every element of $N$. Hence every element of $N$ is real in this case, so it suffices to prove that $|N| \geq [P:\Phi(P)]$.
I am not happy with the previous version of this post, so let me try instead to finish it with a result of Isaacs, (Canadian J. Math, 24, (1972)), which proves that a completely reducible linear group of degree $d$ which is a $p$-group for some prime $p$, can be generated by $\lfloor \frac{3d}{2} \rfloor$ (or fewer) elements. In particular, this implies that if $|N|$ has $d$ prime divisors (including repetitions) ( you probably meant $N$ to be a $q$-group for a fixed prime $q$, but I am allowing the possibility that $|N|$ may have different prime divisors, though all odd), then (Applying the result of Isaacs with $p = 2$), we have $[P: \Phi(P)] \leq 2^{ \frac{3d}{2}} = \sqrt{8}^d < 3^{d} \leq |N|$, as required. Here, we use the faithful action of $P$ on $N$ to view $P$ as a linear group, completely reducible by Mashcke's Theorem as $|N|$ is odd and $P$ is a $2$-group.
For this proof, it has to be noted that $P$ may be viewed as a complex linear group of degree at most $d$: we take the direct sums of the representations afforded by the minimal $P$-invariant subgroups of $N.$ We extend each of these to direct sums of absolutely irreducible representations over finite splitting fields, which does not increase the dimension in any case. We then lift each of these representations to a complex irreducible representation of dimension at most $d,$ where $d$ is the number or prime factors of $|N|$,counting repetitions.
(This problem is more difficult than I thought at first,unless I am just looking at it the wrong way. I would prefer it if someone came up with a more direct proof).
