Let $p: \widetilde{X} \rightarrow X$ form a covering space over $X$. Let
$$ D = \{\sigma : \sigma \text{ is an $\widetilde{X}$ isomorphism}\} $$
form the set of deck transformations of $\widetilde{X}$.
Question 1: If $\sigma \in D$, in what sense is $\sigma$ an isomorphism beyond the fact that $\sigma$ is a bijective mapping?
Question 2: If $\sigma$ happens to fix a point of $\widetilde{X}$, this evidently forces $\sigma = id$ where $id$ is the identity mapping on $\widetilde{X}$. Why is this so?
1) A deck transformation, i.e., an automorphism of a covering space, is a homeomorphism $\sigma : \widetilde{X} \to \widetilde{X}$ such that $p \circ \sigma = p$.
2) Choose a basepoint $x_0 \in X$. Suppose $(Y, y_0)$ is a connected pointed space and $f : (Y, y_0) \to (X,x_0)$ a continuous map of pointed spaces. Given $\widetilde{x_0} \in p^{-1}(x_0)$, then there is at most one map $\tilde{f} : (Y,y_0) \to (\widetilde{X},\widetilde{x_0})$ such that $p \circ \tilde{f} = f$. (Such an $\tilde{f}$ is called a lift of $f$.) This is a standard result in algebraic topology: it's Lemma 10.3 of Rotman or Proposition 1.34 of Hatcher.
Suppose $\sigma$ is a deck transformation that fixes a point $\widetilde{x_0} \in \widetilde{X}$. We can apply the above result to show that $\sigma = \text{id}$. Let $x_0 = p(\widetilde{x_0})$. Then $\sigma$ is a lift of $p : (\widetilde{X}, \widetilde{x_0}) \to (X, x_0)$. But the identity is also a lift of $p$, which implies $\sigma = \text{id}$ by the above result.