Conjectures for finite groups that fail with large counterexamples

According to this answer by Chain Markov on the post I linked to, here are a few, ordered from smallest to largest (known) counterexample:

• Automorphism group of a non-abelian group is non-abelian: counterexample at $N=64$.

• All products of commutators of any finite group are commutators. Counterexample at $N=96$. (Also mentioned in a comment by Gerry Myerson on this question). Indeed, there are $2$ groups of order $96$ for which this fails - those with GAP ID $(96,3)$ and $(96,203)$.

• Automorphism groups of all finite groups not isomorphic to $\{e \}$ or $C_2$ have even order. Counterexample at $N=2187$ (automorphism group of said group has order $729$)

• Moreto conjecture (very similar to the one I put in the body of the question): Let $S$ be a finite simple group and $p$ the largest prime divisor of $|S|$. If $G$ is a finite group with the same number of elements of order $p$ as $S$ and $|G| = |S|$, then $G \cong S$. Also fails at $N=20160$ with the simple groups of that order.

• Any Leinster group has even order. Smallest known counterexample at $N=355433039577$.

• Any nontrivial complete finite group has even order (a conjecture of S. Rose): smallest known counterexample at $N=788953370457$. (c.f. A341298 in the OEIS.)

• Hughes conjecture: suppose $G$ is a finite group and $p$ is a prime number. Then $[G : \langle\{g \in G| g^p \neq e\}\rangle] \in \{1, p, |G|\}$. Smallest known counterexample at $N=142108547152020037174224853515625$.

• Suppose $p$ is a prime. Then, any finite group $G$ with more than $\frac{p-1}{p^2}|G|$ elements of order $p$ has exponent $p$. Smallest known counterexample at $N=142108547152020037174224853515625$ (with $p=5$, same group as in the above one fails).

I'd be interested in any more people have.

Edit: Another answer of mine, kept separate, since it was not on Chain Markov's list.

Another answer from me, separate from my other answer since it was not on the list that I cited there:

Suppose $G$ is the unique group of order $N$ (that is, $G$ is cyclic). Then $N$ has at most $2$ prime factors. Fails first for $N=255 = 3 \times 5 \times 17$.

Interestingly, it'll also fail for all Carmichael numbers: Korselt's criterion for Carmichael numbers happens to be a stricter criterion than the classification of cyclicity-forcing numbers, which can be seen quite directly.

(Less interestingly, $1$ group of order $N$ $\implies N$ prime fails first for $N=15$.)

What about at most $3$ prime factors? Fails first for $N=5865=3 \times 5 \times 17 \times 23$.

At most $4$ prime factors? $N=146965=5 \times 7 \times 13 \times 17 \times 19$.

At most $5$ prime factors? $N=3380195 = 5 \times 7 \times 13 \times 17 \times 19 \times 23$, ...

Yet another answer from me. Not nearly as big, failing at $N=24$ with $2$ groups. For context, there are $59$ groups of order $\leq 23$, $74$ groups of order $\leq 24$.

"The commutator subgroup of a finite group is abelian" fails for two groups of order $24$ (they are $S_4$ and $SL_2(\mathbb{F}_3)$. $[S_4, S_4] = A_4$, $[SL_2(\mathbb{F}_3), SL_2(\mathbb{F}_3)] \cong Q_8$) and no smaller groups.

There's Burnside's conjecture. It's that every class preserving automorphism of a finite group is inner.

He himself eventually discovered a counterexample. I'll have to look it up, but I believe the order was at least $32.$

Stay tuned (wish I could figure out how to use the sandbox).

Yes $32.$ See here. Burnside's counterexample may have been order $64.$