I vaguely remember reading somewhere that a sheaf on a topological space is locally constant if and only if there is an etale map (i.e. a local homeomorphism) such that the inverse image by the etale map is a constant sheaf.
In other words,
Conjecture: A sheaf $\mathscr{L}$ on a space $X$ is locally constant if and only if there is an etale map $p: E \to X$ and a constant sheaf $L$ on $E$ such that $p^{-1} \mathscr{L} \cong L$.
I might be remembering it wrong; it might be that $\mathscr{L}$ is locally constant if and only if for any covering map (locally trivial etale map) $p: E \to X$ that $p^{-1} \mathscr{L}$ is a constant sheaf.
Anyway, if this statement is correct, does anyone happen to know of a reference for it?
This is true, but this is in fact an obvious statement.
If you require only that $p$ is a local homeomorphism, then $p:\coprod U_i\rightarrow X$ works (where the $U_i$ is an open covering which trivializes $\mathcal{L}$).
If you want that $p$ is a covering map, you need some condition on $X$, namely that $X$ has a universal cover. Then $p:\tilde{X}\rightarrow X$ works (every locally constant sheaf on a simply connected space is constant).
Of course, you cannot replace "there exist" by "for all" as you seem to suggest.
Conversely, if there exist a local homeomorphism $p:Y\rightarrow X$ (for example a covering map), then $Y$ is covered by open subset $V_i$ such that $p:V_i\overset{\sim}\rightarrow U_i$ is an homeomorphism. Thus, $\mathcal{L}$ is constant on the $U_i$'s (because its inverse images on the $V_i$'s are constant) so $\mathcal{L}$ is locally constant.