# Galois Group as Semi-Direct Product

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 $$\sigma \in \text{Gal}(K_{n}/K)$$?