What is known about the number of groups of order $p^k$ with $|Z|=p$ , which I denote $N(p^k)$ ?
For $k=1$ , it is clear, that we have $N(p^k)=1$ : We have one group of order $p$ and since it is abelian, we also have $|Z|=p$.
For $k=2$ , we have $N(p^k)=0$ : We have two groups of order $p^2$, both are abelian, so $|Z|=p$ is impossible.
For $k=3$, we have $N(p^k)=2$ : There are $3$ abelian and $2$ non-abelian groups. Since $|Z|=p^2$ is impossible, the claim follows.
For $k=4$ , I conjecture that for $p>2$ we have $N(p^k)=4$. For $p=2$, we have $N(p^k)=3$. GAP approves this for $p\le 1000$.
For $k=5$ , I also checked the primes upto $1000$ and noticed that $N(p^k)$ is one of the numbers $10,14,15,17,19,21$. So, I conjecture this is the case for every $p$.
Is there a proof for the two conjectures ?
Can $N(p^k)$ be estimated for larger $k$ ?
Does $$\lim_{k\rightarrow \infty}\frac{N(p^k)}{gnu(p^k)}=0$$ hold for every $p$ , where $gnu(p^k)$ denotes the number of groups of order $p^k$ ?
Groups of order $p^4$ and $p^5$ are classified, so it should be a matter of chasing references to check your conjecture. The following is a good starting point:
Besche, Hans Ulrich; Eick, Bettina; O'Brien, E. A. A millennium project: constructing small groups. Internat. J. Algebra Comput. 12 (2002) 623–644.
EDIT: As for your second conjecture, I'm not so sure anymore.