descends to an automorphism of $G(r,p,q,n)$ which we denote by the same symbol. Note that $\tau^2 = 1$.

Related to this automorphism is the following fact, proved by the first author in [@Ca2]. Let ${\mathrm{Irr}}(G)$ denote the set of complex irreducible characters of a finite group $G$, and fix positive integers $r,p,q,n$ with $p$ and $q$ dividing $r$ and $pq$ dividing $rn$.

\[f\] Let $\tau$ denote the inverse transpose automorphism of $G=G(r,p,q,n)$ defined above. Then $$|\cI_{G,\tau}| {\stackrel{\rm def}{=}}\left|\left\{ \omega \in G : \omega^{-1} = \tau(\omega)  \right\} \right| \leq \sum_{\psi \in {\mathrm{Irr}}(G)} \psi(1).$$ Furthermore, equality holds if and only if (i) $\gcd(p,n) \leq 2$ or (ii) $\gcd(p,n) = 4$ and $r\equiv p \equiv q \equiv n \equiv 4 \modu 8)$.

A *class-preserving outer automorphism* of a group $G$ is an automorphism which sends every element to a conjugate element but which is not a map of the form $g\mapsto xgx^{-1}$ for some $x \in G$. The following lemma is immediate from the preceding theorem and [@Ma Lemma 5.1].

\[5.1sub\] Suppose $G=G(r,p,q,n)$ has no class-preserving outer automorphisms and equality holds in Theorem \[f\]. Then $G$ has a GIM if and only if $G$ has a GIM with respect to the automorphism $\tau 