Let $G$ be a finite group and $f$ be an automorphism of $G$. We say that $f$ has a regular orbit if there exists $x\in G$ such that $|x^f|=|f|$. If $G$ is abelian it is known that every automorphism of $G$ has a regular orbit.
If $L$ is a finite simple non-abelian group, can I always find an automorphism of $L$ with no regular orbit?
By a result of Horoševskiĭ you can never find such an automorphism, that is all automorphisms of finite simple groups have a regular orbit.
As pointed out by Michael Giudici the answer is given by a result of Horoševskiĭ. Here is a proof following the paper by Horoševskiĭ.
Lemma: Let $\phi$ be an automorphism of $G$ with $|\phi|$ divisible by $p^2$ for some prime $p$. Then $\phi$ has a regular orbit if and only if $\phi^p$ has a regular orbit.
Proof: If the orbit of $\phi$ on some $x \in G$ is regular, clearly the orbit of $\phi^p$ on $x$ is also regular. Conversely suppose that the orbit of $\phi^p$ on $x \in G$ is regular, so $C_{\langle \phi^p \rangle}(x) = 1$. We have $C_{\langle \phi \rangle}(x) = \langle \phi^r \rangle$ for some $r$. Since $\langle \phi^r \rangle \cap \langle \phi^p \rangle = 1$ and the order of $\phi^p$ is divisible by $p$, we have $p \nmid |\phi^r|$. But then $p$ must divide $r$, so $\langle \phi^r \rangle = 1$.
Theorem: Let $G$ be a finite simple group and $\phi$ an automorphism. Then $\phi$ has a regular orbit on $G$.
Proof: By the lemma, we can assume that $|\phi| = N = p_1 \cdots p_n$ with $p_1 < \cdots < p_n$ distinct primes. The result is obvious for $n = 1$, so suppose $n > 1$ and proceed by induction on $n$.
For contradiction suppose that $\phi$ has no regular orbit on $G$, in which case $$G = \bigcup_{i = 1}^n C_G\left(\phi^{N/p_i}\right).$$
Here $G$ is a union of $n$ subgroups, so one of these subgroups must have index $< n$. (If $G = \cup_{i = 1}^n H_i$, then $|G| < \sum_{i = 1}^n |H_i|$, so for $|H_j|$ maximal $|G|/|H_j| < n$.)
Then because $G$ is simple, we get an embedding of $G$ into the symmetric group of degree $n-1$, so $|G| \le (n-1)!$. On the other hand by induction $\phi^{p_1}$ has a regular orbit on $G$, so $|G| > |\phi^{p_1}| = p_2 \cdots p_n$. Here $p_2 \cdots p_n > (n-1)!$ since the $p_i$ are distinct primes, so we have a contradiction.
If I am reading your question correctly, then I think $A_{5}$ is an example where this fails. The automorphism group is isomorphic to $S_{5}$. The only elements of composite order in the automorphism group have order $6$ and $4$. If $\sigma$ is an automorphism of order $6$, then $\sigma^{2}$ has $3$ fixed points (by conjugation on the simple group) and $\sigma^{3}$ has $6$ fixed points, while $\sigma$ has $9$ regular orbits. If $\tau$ is an automorphism of order $4$, then $\tau^{2}$ has $4$ fixed points, so $\tau$ has $14$ regular orbits. An automorphism of prime order $p$ has at most $6$ fixed points, so has many regular orbits. Note that $A_{5} \cong {\rm SL}(2,4),$ and I think there will be other cases where ${\rm SL}(2,2^{p})$ will not have this property with $p > 3$ a prime, but I have not checked (the case where $2^{p}-1$ is a Mersenne prime seems even more promising). It seems likely to me that it is actually rather rare to find an automorphism of a finite non-Abelian simple group which has no regular orbit.
