If $|G|=p^3q^2$ then $\Phi(G)$ is cyclic for primes $p\neq q$. I have conjectured this result for the Frattini subgroup by doing some calculations in GAP. I think this is even true if $|G|=p_1^{i_1}\cdots p_n^{i_n}$ for $i_j\leq 3$ holds, but I would like to stick with this example first. I know the following:
If $p\mid |G|$, then $p\mid |G/\Phi(G)|$, thus the order of $\Phi(G)$ is at most $p^2q$. Since $\Phi(G)$ is nilpotent, it must be abelian, too.
That's why I think that it suffices to show this result for only abelian $G$ (which is fairly easy), as considering all abelian groups with that order would lead to all possible abelian $\Phi(G)$. But I still need a valid argument for that.
I also tried the assumption $\Phi(G)=C_p\times C_p$ or $\Phi(G)=C_p\times C_p\times C_q$ but I have not yet found a contradiction.
I would be glad to see some ideas or approaches for this conjecture.
 A: Actually, the generalised version of your conjecture that says "If $|G|=p_1^{i_1}\cdots p_n^{i_n}$ for $i_j\leq 3$, then $\Phi(G)$ is cyclic" is true.
To prove it we will first prove three small lemmas:
Lemma 1
Suppose $G$ is a finite group and $H$ is its proper subgroup. Suppose $\phi: G \rightarrow K$ is a homomorphism, such, that $Ker \phi \leq H$. Then $H$ is maximal proper subgroup in $G$ iff $\phi(H)$ is maximal proper subgroup of $\phi(G)$.
Proof:
If $H < J < G$, then $[\phi(J):\phi(H)] = [J:H] > 1$ and $[\phi(G):\phi(J)] = [G:J] > 1$, that implies $\phi(H) < \phi(J) < \phi(G)$.
If $\phi(H) < \phi(J) < \phi(G)$, then $[J:H] = [\phi(J):\phi(H)] > 1$ and $[G:J] = [\phi(G):\phi(J)] > 1$, that implies $H < J < G$.
Thus $H$ is maximal proper subgroup in $G$ iff $\phi(H)$ is maximal proper subgroup of $\phi(G)$ Q.E.D
Lemma 2
Suppose $G$ is a finite group and $\phi: G \rightarrow K$ is a homomorphism, such, that $Ker \phi \leq \Phi(G)$. Then $\Phi(\phi(G)) = \phi(\Phi(G))$.
Proof:
Suppose $\{H_i\}_{i = 1}^{n}$ are all maximal proper subgroups of $G$. Then, by Lemma 1, all maximal proper subgroups of $\phi(G)$ are exactly $\{\phi(H_i)\}_{i = 1}^{n}$. Thus $\Phi(\phi(G)) = \bigcap_{i = 1}^n \phi(H_i) = \phi(\bigcap_{i = 1}^n H_i) = \phi(\Phi(G))$ Q.E.D.
Lemma 3
Suppose $G$ is a finite group and $\pi$ is a set of primes. Then $\Phi(G)$ has a unique Hall $\pi$-subgroup and it is characteristic in $G$.
Proof:
As $G$ is finite, $\Phi(G)$ is nilpotent. Thus it is a direct product of its Sylow subgroups. So, all its Hall subgroups are unique and characteristic in $\Phi(G)$. So they are characteristic in $G$ too, as $\Phi(G)$ is characteristic in $G$.
Now, let’s prove your conjecture.
As $\Phi(G)$ is nilpotent, it is cyclic iff all its Sylow subgroups are cyclic. Now, suppose $p$ is a prime divisor of $|G|$. Suppose, $P$ is the unique Sylow $p$-subgroup of  $\Phi(G)$. Then, $|P|$ is either $p$ or $p^2$. Suppose, $P$ is non-cyclic. Then it is isomorphic to $C_p \times C_p$. Suppose $Q$ is the Hall subgroup of $\phi(G)$, corresponding to the set of all prime divisors of $G$ except $p$. Then you can see, that $P \times Q = \phi(G)$. By Lemma 3, $Q$ is characteristic and thus normal. Now, let’s consider the group $\frac{G}{Q}$. $|\frac{G}{Q}| | |G|$ and thus $|\frac{G}{Q}|=p_1^{i_1}\cdots p_n^{i’_n}$ for $i’_j\leq 3$ And by Lemma 2, $\Phi(\frac{G}{Q}) = \frac{\Phi(G)}{Q} = P$. However, for every finite group $H$, $\Phi(G) \cong C_p \times C_p$ implies $p^4 | |H|$ (the proof of this fact can be found here: A question about Frattini subgroup of specific form). So we get a contradiction. That means, that $\Phi(G)$ is indeed cyclic. Q.E.D.
