Let $\Gamma$ be a finite group that contains an involution. We assume that the cardinality of the first cohomology group $H^1(\mathbb{Z}/2\mathbb{Z}, \Gamma)$ is greater than the size of the conjugacy classes of $\Gamma$, denoted as $\mathrm{conj}(\Gamma)$. That is, we have: \[ \vert H^1(\mathbb{Z}/2\mathbb{Z}, \Gamma) \vert > \vert \mathrm{conj}(\Gamma) \vert. \]

No.

A class in $H^1(G, \Gamma)$ corresponds to an element $g \in \Gamma$ satisfying the condition $g \sigma(g) = 1$, modulo an equivalence relation defined by $g_1 \sim g_2$ if there exists some $h \in \Gamma$ such that $g_1 = h g_2 \sigma(h)^{-1}$. This equivalence relation is known as $\sigma$-conjugacy.

We have the following chain of inequalities, where $(g)$ represents the (ordinary) conjugacy class of $g$: $$ |H^1(G,\Gamma)| \leq | \Gamma/ \sigma\textrm{-conjugacy}| = | \{ (g) \in \operatorname{conj}(\Gamma) \mid (\sigma(g))=(g) \} | \leq | \operatorname{conj}(\Gamma)| $$ These inequalities collectively lead to a negative conclusion regarding your question. The validity of both inequalities is straightforward, so my attention will be directed towards demonstrating the equality.

The orbit-stabilizer theorem indicates that when we take the sum of the sizes of the stabilizers for each element in an orbit under a $\Gamma$-action, the result equals the order of the group $\Gamma$. When we examine the action of $\Gamma$ on itself through $\sigma$-conjugacy, this leads us to conclude that

We have the equation $$|\Gamma| |\Gamma/ \sigma\textrm{-conjugacy} | = | \{ g, h \in \Gamma \mid g = h g \sigma(h)^{-1} \}|.$$ This holds because each element \( g \) in a given \( \sigma \)-conjugacy class contributes \( |\Gamma| \) to the count on the right side. The condition \( g = h g \sigma(h)^{-1} \) is equivalent to saying that \( \sigma(h) = g^{-1} h g \). Thus, we can reformulate the relationship between \( g \) and \( h \) in terms of the action of \( \sigma \) on \( h \).

$$| \{ g, h \in \Gamma \mid g = h g \sigma(h)^{-1} \}| = | \{ g, h \in \Gamma \mid \sigma(h) = g^{-1} h g \}|.$$

The element $h$ adds to this total if and only if $\sigma(h)$ is conjugate to $h$. When it does contribute, the count of solutions corresponds to the centralizer of $h$, which serves as the stabilizer for the conjugation action. Consequently, the contribution to the set $| \{ g, h \in \Gamma \mid \sigma(h)=g^{-1}h g\}|$ from $h$ within each conjugacy class equals $|\Gamma|$ if the conjugacy class remains invariant under $\sigma$, and it is $0$ in other cases. Therefore,

To verify the claimed identity, we start by analyzing both sides of the equation: