= Solution
Choose a path $\gamma$ in $X$ from $x_0$ to $x_1$. For every $\widehat x\in p^{-1}(x_0)$, the <path lifting theorem> gives a unique lift of $\gamma$ starting at $\widehat x$. Send $\widehat x$ to the endpoint of this lift. Lifting the reversed path gives the inverse map, so this is the <fibre bijection by path lifting>
$$
p^{-1}(x_0)\cong p^{-1}(x_1).
$$
In particular, all fibres have the same cardinality.
A connected covering is a <normal covering map> when its deck transformations act transitively on a fibre. Equivalently, for a choice of $\widehat x_0$ above $x_0$,
$$
p_*\pi_1(\widehat X,\widehat x_0)
\triangleleft
\pi_1(X,x_0).
$$
The relevant <lifting criterion for a covering space> says that a based map $f:(Y,y_0)\to(X,x_0)$ lifts through $p$ exactly when
$$
f_*\pi_1(Y,y_0)
\subseteq
p_*\pi_1(\widehat X,\widehat x_0).
$$
For a universal covering, $\widehat X$ is simply connected, so the displayed covering subgroup is trivial and hence normal. Thus a <universal covering map is normal>.
Now consider connected finite covers of the closed orientable surface $\Sigma_g$. By the <classification of connected covering spaces>, degree-$n$ connected covers correspond to index-$n$ subgroups of the <fundamental group of a closed orientable surface>, and normal covers correspond to normal subgroups.
The cases in which normality is forced are:
* $n=1$, because the subgroup is the whole fundamental group.
* $n=2$, because every <index-two subgroup is normal>.
* $g=1$, because $\pi_1(\Sigma_1)\cong\mathbb Z^2$ is abelian, so all its subgroups are normal.
* For $g=0$, the sphere is simply connected, so a connected cover necessarily has $n=1$.
It remains to show that these are the only forced cases. Let $g\geq2$ and $n\geq3$. Write
$$
\pi_1(\Sigma_g)=
\left\langle a_1,b_1,\ldots,a_g,b_g
\mathrel{\Big|}
\prod_{i=1}^g[a_i,b_i]=1
\right\rangle.
$$
In the <symmetric group> $S_n$, put
$$
\sigma=(1\,2\,\cdots\,n),
\qquad
\tau=(1\,2).
$$
Define a homomorphism by
$$
a_1\mapsto\sigma,\quad b_1\mapsto\tau,\quad
a_2\mapsto\tau,\quad b_2\mapsto\sigma,
$$
and send all remaining generators to the identity. The relation is respected because
$$
[\sigma,\tau][\tau,\sigma]=1.
$$
The cycle $\sigma$ and transposition $\tau$ generate $S_n$, so the homomorphism is surjective.
Let
$$
H=\phi^{-1}\bigl(\operatorname{Stab}_{S_n}(1)\bigr).
$$
The natural action of $S_n$ is transitive, so $H$ has index $n$. Its image is the point stabilizer $S_{n-1}$, which is not normal in $S_n$ for $n\geq3$; hence $H$ is not normal. The connected covering corresponding to $H$ is therefore an explicit degree-$n$ nonnormal cover. This is the <nonnormal finite cover of a higher-genus orientable surface>.
Consequently, among connected covers that exist, normality is forced exactly when
$$
\boxed{n\leq2\quad\text{or}\quad g\leq1,}
$$
with the qualification that $g=0$ permits only $n=1$, as summarized by the <forced normality of finite connected covers of orientable surfaces>.
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