This is with regards to this lemma.
Lemma: Let $p: \tilde{X} \rightarrow X$ be a covering map. Assume $\tilde{X}$ is connected and locally path-connected. Then $p$ is a regular covering $\iff$ for every pair of $\tilde{x_1}$ and $\tilde{x_2}$ such that $p(\tilde{x_1})=p(\tilde{x_2})$, there exists a deck transformation $h$ such that $h(\tilde{x_1})=\tilde{x_2}$.
($\Leftarrow$) Consider $\tilde{x_1},\ \tilde{x_2} \in \tilde{X}$ such that $p(\tilde{x_1})=p(\tilde{x_2})$.
Let $h \in G(\tilde{X},p,X)$ (deck transformation) such that $h(\tilde{x_1})=\tilde{x_2}$.
Then $p_*(\pi_1(\tilde{X},x_1))=p_*(h_*(\pi_1(\tilde{X},x_1)))=p_*(\pi_1(\tilde{X},x_2))$
Since $p_*(\pi_1(\tilde{X},x_1))$ and $p_*(\pi_1(\tilde{X},x_2))$ are related by conjugation, we have that $p$ is regular. $\leftarrow$
The left arrow above denotes the place where I have difficulty understanding. I understand that there is only one subgroup in the conjugacy class (the subgroups of $\pi_1(X,x)$ related by conjugation), but how does that imply that $p$ is regular?
Please kindly avoid using too complicated abstract algebra concepts; I have only basic knowledge of groups and rings.
Denote $x=p(\tilde x_1)=p(\tilde x_2)$. The point is to show that $p_*\pi_1(\tilde X,\tilde x_1)$ is normal in $\pi_1(X, x)$ which amounts to show that the normalizer subgroup of $p_*\pi_1(\tilde X,\tilde x_1)$ in $\pi_1(X,x)$ is the whole group $\pi_1(X,x)$. So let $g$ be any element in $\pi_1(X,x)$ we claim that $g$ is in the normalizer of $p_*\pi_1(\tilde X,\tilde x_1)$:
The element $g$ represents a loop $\gamma$ based in $x$. The lift of $\gamma$ from $\tilde x_1$ is a path $\tilde \gamma$ starting at $\tilde x_1$ and ending at some point $\tilde x_2$ belonging to the fiber $p^{-1}(x)$. It can be shown that going from one point of the fiber to another amounts to conjugate the subgroups $p_*\pi_1(\tilde X,\tilde x_i)$. That is, $$p_*\pi_1(\tilde X,\tilde x_2)=g\;p_*\pi_1(\tilde X,\tilde x_1)\;g^{-1}\;\; (1)$$ The existence of a Deck transformation taking $\tilde x_1$ to $\tilde x_2$ imposes that $$p_*\pi_1(\tilde X,\tilde x_2)=p_*\pi_1(\tilde X,\tilde x_1)\;\;(2)$$ combining $(2)$ with $(1)$ gives that $$p_*\pi_1(\tilde X,\tilde x_1)=g\;p_*\pi_1(\tilde X,\tilde x_1)\;g^{-1}\;\; (3)$$ Which means that $g$ is in the normalizer of $p_*\pi_1(\tilde X,\tilde x_1)$, and the claim follows.