Let $p$ be a prime and $X$ be finite graph with its automorphism group ${\mathrm{Aut}}(X)$ cyclic of order $p^e \geq 7$.
If the number of vertices $|V(X)| < 2p^e$, the ${\mathrm{\bf claim}}$ is for any $\varphi \in {\mathrm{Aut}}(X)$ and any $y \in V(X)$, either $\varphi(y) = y$, or else the vertices $y, \varphi(y), \dotsc, \varphi^{p^e - 1}(y)$ are distinct (Ref. On the minimum order of graphs with given Automorphism group, Sabidussi, 1958). As a consequence ${\mathrm{Aut}}(X)$ must be dihedral of order $2p^e$ and hence $|V(X)| \geq 2p^e$.
I am unable draw the consequence from the claim. In fact I am not sure about the claim itself. From class equation, any non-trivial orbit is a factor of $p^e$, but not necessarily equal.
I could prove rather the following: if ${\mathrm{Aut}}(X)$ cyclic of order $p^e$ and $\varphi$ is a generator, then there exists an $y \in V(X)$ such that $y, \varphi(y), \dotsc, \varphi^{p^e - 1}(y)$ are distinct.
I can't follow this claim of Sabidussi either. In fact, I did a quick search on this problem, and it seems the claim is wrong, and the correct bound is $p^e+p$ for graphs with automorphism group cyclic of prime power order $p^e\geq 7$.
See for example:
Smallest graphs with given automorphism group, Danai Deligeorgaki, Journal of Algebraic Combinatorics 56, 609–633 (2022)