Is there a surjective local homeomorphism which is "normal" but not a covering map? Let $X\xrightarrow{Q}Y$ be a surjective local homeomorphism, between topological spaces.
Suppose $Q$ is "normal", in that for every $x,x' \in X$, if $Q(x) = Q(x')$, then there exists a self-homeomorphism $f \in\mathrm{Homeo}(X)$
such that $f(x) = x'$ and $Q \circ f = Q$.
Then, can $Q$ still fail to be a covering map?
 A: Sure.  Suppose $X$ is the line with two origins.  Concretely, $X$ is the quotient of $\mathbb{R}\times \{-1,1\}$ by the equivalence relation which identifies $(t,-1)$ with $(t,1)$ for any non-zero $t$, but it does not identify $(0,-1)$ with $(0,1)$.  I'll refer to points in $X$ as real numbers, except there are two origins $0_A$ and $0_B$.
Take $Y$ to be the usual real line $\mathbb{R}$.  Let $Q:X\rightarrow Y$ be the function $Q(t) = t$, $Q(0_A) = Q(0_B) = 0$.  Using the definition of quotient topology, it's easy to see that $Q$ is a local homeomorphism.
Moreover, $Q$ is injective except that $Q(0_A) = Q(0_B)$.  But is again easy to see that the homeomorphism $f:X\rightarrow X$ given by $f(t) = t$, $f(0_A) = 0_B$, $f(0_B) = 0_A$ satisfies $Q\circ f = Q$.
A: Examples arise naturally as the étalé spaces of sheaves, using a group structure on the sheaf to obtained the required automorphisms.  For instance, let $Y=\mathbb{R}$ and let $Q:X\to \mathbb{R}$ be the étalé space of the sheaf of continuous real-valued functions on $\mathbb{R}$.  This is "normal" in your sense because any element of $X$ (i.e., a germ of a continuous function at a point of $\mathbb{R}$) can be extended to a global section, and then by subtracting such a global section you see that automorphisms of the sheaf act transitively on each stalk.  However, it is not a covering map, for instance because $X$ is not Hausdorff (consider two functions that agree on $[0,\infty)$ but disagree on all of $(-\infty,0)$; then their germs at $0$ do not have disjoint neighborhoods).
(More generally, the étalé space of any soft sheaf of groups will be "normal", but will rarely be a covering map.)
