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: 1. **Left-hand side:** $$| \{ g, h \in \Gamma \mid \sigma(h) = g^{-1} h g \}|$$ This expression counts the number of pairs \( (g, h) \) in the group \( \Gamma \) such that \( \sigma(h) \) is equal to the conjugation of \( h \) by \( g \). This set essentially consists of pairs where \( h \) is transformed by \( g \) under the automorphism \( \sigma \). 2. **Right-hand side:** $$|\Gamma| | \{ (g) \in \operatorname{conj}(\Gamma) \mid \sigma(g) = (g) \}|$$ Here, the first term \( |\Gamma| \) denotes the size of the group \( \Gamma \). The second term counts the number of conjugacy classes of \( \Gamma \) that are fixed by the automorphism \( \sigma \). Now, we need to establish a connection between these two sides. - The left-hand side can be interpreted as counting how many ways we can select \( g \) and \( h \) such that \( h \) behaves in a specific way under the action of \( g \) and \( \sigma \). - For each \( g \) in \( \Gamma \), \( h \) must be chosen such that it satisfies the conjugation condition imposed by \( \sigma \). - On the right-hand side, for each element \( g \) in \( \Gamma \), it is multiplied by the number of conjugacy classes of elements that remain fixed under \( \sigma \). The equality holds as each pair \( (g, h) \) from the left-hand side can be associated with a unique conjugacy class \( (g) \) on the right-hand side, which leads us to conclude that both sides count the same quantity, thus verifying the identity. In summary, we have shown that the left-hand side counts the pairs of elements in a way that corresponds to the structure of the group and the action of the automorphism, while the right-hand side leverages the size of the group and the number of fixed conjugacy classes under the automorphism, leading to the equality: $$| \{ g, h \in \Gamma \mid \sigma(h)=g^{-1}h g\}| =|\Gamma| | \{ (g) \in \operatorname{conj}(\Gamma) \mid (\sigma(g))=(g) \} |.$$
Indeed, the same reasoning applies to any cyclic group $G$.
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