Is $A_5$ the only finite simple group with only 4 distinct sizes of orbits under the action of the automorphism group? Given a finite group $G$, let $\eta(G)$ denote the number of distinct
sizes of orbits on $G$ under the action of ${\rm Aut}(G)$.
It happens that there are infinitely many non-abelian finite simple groups
$G$ such that $\eta(G) = 5$. For example, this holds for the groups
${\rm PSL}(2,p)$ for prime numbers $p \geq 7$.
In contrast, is it true that $G = {\rm A}_5$ is the only non-abelian finite
simple group such that $\eta(G) = 4$? --
GAP computations suggest this may be true.
Remark: I know that among the alternating groups, the sporadic groups and
the simple groups of Lie type except for those of type $^2{\rm A}$, ${\rm A}$, ${\rm B}$,
${\rm C}$, ${\rm D}$, $^2{\rm D}$ and $^3{\rm D}_4$, there are no further
groups $G$ with $\eta(G) \leq 4$.
Update (April 30, 2020): Meanwhile this question has 3 deleted answers.
 A: Let me see if I can write down a proof that the answer is YES for most of the $A_n$-groups. The method should work for the rest of the $A_n$-groups and, indeed, all of the other groups you mention. Specifically, I'll prove
Proposition: If $G={\rm PSL}_n(q)$ with $n\geq 5$ and $q\geq 2$, then $\mu(G)>4$.
Proof: In what follows, I'll write $q=p^f$ where $p$ is a prime, $f$ a positive integer. Then $|{\rm Aut}(G)|/ |{\rm PGL}_n(q)|=2f$. Recall that a primitive prime divisor of $p^{df}-1$ is a prime that divides $p^{df}-1$ but not $p^k-1$ for any integer $k<df$. Zsigmondy's theorem asserts that such a prime always exists unless $df=2$ or $(p,df)=(2,6)$.
An easy argument shows that if $r$ is a primitive prime divisor of $p^{df}-1$, then $r>df$. It's written down as Lemma 2.7 of a paper of mine with Azad and Britnell. This is important because it means that primitive prime divisors of $p^{df}-1$ do not divide $2f$ provided $df>1$.
So now our job is to find five elements, $g_1,\dots, g_5\in G$, with different orbit sizes under ${\rm Aut}(G)$. Write $o(g_i)$ for the the orbit size of $g_i$. In what follows we note down what makes each $o(g_i)$ definitely different to the others.


*

*Let $g_1=1$. Then $o(g_1)=1$.

*Let $g_2$ be central in a Sylow $p$-subgroup of $G$. Then $o(g_i)$ is not divisible by $p$.

*Let $g_3$ be an element whose centralizer is a maximal torus in ${\rm PGL}_n(q)$ of size $\frac{q^n-1}{q-1}$. Then $o(g_3)$ is not divisible by a ppd of $q^n-1$.

*Let $g_4$ be an element whose centralizer is a maximal torus in ${\rm PGL}_n(q)$ of size $q^{n-1}-1$. Then $o(g_4)$ is not divisible by a ppd of $q^{n-1}-1$.

*Let $g_5$ be an element whose centralizer is a maximal torus in ${\rm PGL}_n(q)$ isomorphic to $(q^{n-2}-1)\times(q-1)$. Then $o(g_5)$ is not divisible by a ppd of $q^{n-2}-1$.


Some remarks:


*

*the existence of tori of this size can be seen directly, but is written down explicitly in the paper of Buturlakin and Grechkoseeva.

*The fact that a ppd of $q^{n-2}-1$ does not divide $q^n-1$ follows from the fact that $n\geq 5$ and the fact that $gcd(q^{n-2}-1, q^n-1)$ divides $q^2-1$. Similarly for the other pairs of ppds.

*We are implicitly using the lower bound on ppd's used above -- so that the action of $|{\rm Out}(G)|$ doesn't mess things up.

*In theory one should check what happens when $(q,n)\in\{(2,6), (2,7), (2,8)\}$ -- here Zsigmondy's theorem fails for one of the tori mentioned above and a ppd does not exist. But I've excluded $q=2$ so this doesn't arise (see next comment).

*Finally the fact that there really are elements that are centralized by these maximal tori is a fairly straightforward eigenvalue argument. The only problem arises when $q=2$ and we have the torus $(q^{n-2}-1)\times (q-1)$, hence I've excluded $q=2$ from the statement of the theorem. QED


Extra remarks


*

*Dealing with $q=2$ should be easy. For instance if $n$ is odd, then you can substitute that torus of size $(q^{n-2}-1)\times (q-1)$ with one of size $(q^{n-2}-1)\times (q+1)$ (maybe this will work for $n$ even too but I haven't checked).

*Likewise $n=2,3,4$ can be done by hand (I guess).

*The whole argument carries over to the unitary groups just by changing a few signs in the size of the tori. The other classical groups will require that you choose tori of different sizes.

