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I am trying to understand the following argument but probably have some stupid misunderstanding.

Let $ K = \mathbb{Q} ( \sqrt{-D} ) $ be an imaginary quadratic field, and let $ K_{n} $ be the ring class field of conductor $ n $ where $ n $ is square free. We have the following exact sequence $$ 1 \to \text{Gal}(K_{n} / K ) \to \text{Gal}( K_{n} / \mathbb{Q} ) \to \text{Gal} ( K / \mathbb{Q} ) . $$ The group $ \text{Gal}( K / \mathbb{Q} ) = \left \{ 1 , \tau \right \} $ where $ \tau $ is the complex conjugation. We can lift $ \tau $ to an involution $ \tilde{\tau} $ in $ K_{n} $ such that it acts on the normal subgroup $ \text{Gal}(K_{n} / K ) $ by $ \sigma \mapsto \tilde{\tau} \sigma \tilde{ \tau } ^{-1} = \sigma^{-1} $ and hence $$ \text{Gal}(K_{n} / \mathbb{Q} ) \cong \text{Gal} ( K_{n} / K ) \rtimes \text{Gal}(K/\mathbb{Q} ) . $$

I have two questions.

1) Why can I say that $ \tilde{\tau} \sigma \tilde{\tau} ^{-1} = \sigma ^{-1} $ for all $ \sigma \in \text{Gal}(K_{n} / K ) $? I can only conclude that $ \tau \sigma \tau^{-1} \in \text{Gal}(K_{n} / K ) $.

2) Is any other lift $\eta $ of $ \tau $ an involution and satisfies the same relation? More precisely, I know that I can take $ \tilde{\tau} $ to be the complex conjugation on $ K_{n} $ and then any other $ \eta = \tilde{\tau} \delta $ for $ \delta \in \text{Gal}(K_{n}/K ) $. Would this lift also satisfy $ \eta \sigma \eta^{-1} = \sigma^{-1} $ for all $ \sigma \in \text{Gal}(K_{n}/K) $?

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