Let $q=p^f$ be a power of the prime $p$ and let $\alpha$ be an inner diagonal automorphism of ${^2}{\operatorname{E}_6(q^2)}$, i.e., an element of $({^2}\operatorname{E}_6)_{\mathrm{ad}}(q^2)$ in the notation of [1, pp. 39ff.].

Question: Is $|\{x\in{^2}{\operatorname{E}_6(q^2)}\mid\alpha(x)=x\}|=|\operatorname{C}_{{^2}{\operatorname{E}_6(q^2)}}(\alpha)|\geq q^D$ for some explicit, absolute constant $D>0$?

Some remarks on this Question:

- When replacing ${^2}{\operatorname{E}_6(q^2)}$ by any other finite simple group of Lie type ${^t}{X_r(q^t)}$, the answer to the above Question is "yes" by known results:
- For simple Chevalley groups and simple classical groups, this follows from [2, Theorem 6.15] (the value of the universal constant $A$ mentioned there can be worked out from the bounds in the single cases proved in [2, Section 6]).
- All finite simple groups of Lie type ${^t}{X_r(q^t)}$ that are neither Chevalley groups nor classical groups nor ${^2}{\operatorname{E}_6(q^2)}$ do not have (nontrivial) diagonal automorphisms, and so for these, the problem is about centralizers of inner automorphisms $\alpha$ only. When $\alpha$ is semisimple (i.e., corresponds to a $p'$-element of ${^t}{X_r(q^t)}$), then $\operatorname{C}_{{^t}{X_r(q^t)}}(\alpha)$ contains a maximal torus of ${^t}{X_r(q^t)}$, which by [3, Lemma 3.3] has size at least $\frac{1}{|\operatorname{Out}({^t}{X_r(q^t)})|}\cdot(q-1)^r\geq q^{1/2}$ except possibly for some small $q$ (up to an explicitly determinable upper bound), which can be ignored. On the other hand, when $p\mid\operatorname{ord}(\alpha)$, then [4, proof of Theorem A1, p. 320] gives $|\operatorname{C}_{{^t}{X_r(q^t)}}(\alpha)|\geq q$.

- The "torus argument" for semisimple elements of ${^t}{X_r(q^t)}$ above also works for semisimple inner diagonal automorphisms of ${^2}{\operatorname{E}_6(q^2)}$, so one can assume that $p\mid\operatorname{ord}(\alpha)$ in the above Question.
- Possibly, the argument given in [4, proof of Theorem A1, p. 320] to show that centralizers of non semisimple elements of ${^t}{X_r(q^t)}$ contain some root subgroup over some extension field of $\mathbb{F}_q$ (and thus have order at least $q$) can be extended to non semisimple elements of $({^2}\operatorname{E}_6)_{\mathrm{ad}}(q^2)$, but this is not obvious to me.

References:

[1] R.W. Carter, *Finite groups of Lie type. Conjugacy classes and complex characters*, John Wiley & Sons (Chichester), 1985.

[2] J. Fulman and R. Guralnick, Bounds on the number and sizes of conjugacy classes in finite Chevalley groups with applications to derangements, *Trans. Amer. Math. Soc.* **364**(6):3023-3070 (2012).

[3] B. Hartley, A general Brauer-Fowler theorem and centralizers in locally finite groups, *Pacific J. Math.* **152**(1):101-117 (1992).

[4] B. Hartley and M. Kuzucuoğlu, Centralizers of elements in locally finite simple groups, *Proc. London Math. Soc. (3)* **62**(2):301-324 (1991).