Finding tetration in a multiplicative group modulo p I have a variant on the discrete logarithm problem, involving finding tetration in a multiplicative cyclic group of integers modulo a large prime $p$:

$$a = x^x \mod p$$

Where $a$ and $p$ are known, and $p$ is not necessarily a safe prime. Can $x$ efficiently be found or is this at least as hard as the DLP?
As pointed out in the comments, it's not quite clear what the problem is, but here's an easy algorithm for the (arguably) most natural interpretation.

Given any prime $p$ and integer $a$, the following procedure finds an integer $x\in\mathbb Z_{\geq0}$ such that
$$ x^x\equiv a\pmod p \,\text. $$




Pick some positive integer $e$ coprime to $p-1$. (For example, $e=1$ always works.)
Write the prospective solution as $x=e+k(p{-}1)$ with $k\in\mathbb Z_{\geq0}$.
Since the order of $x$ must be a divisor of $p-1$, the equation becomes
$$(e+k(p{-}1))^{e}\bmod p=a \,\text. $$
Raise everything to the power of $f:=e^{-1}\bmod(p{-}1)$ to obtain
$$ e+k(p{-}1) \equiv a^f \pmod p \,\text. $$
Simply solve this congruence for $k$, that is, compute
$$ k := (e-a^f)\bmod p \,\text. $$
Output $x:=e+k(p{-}1)$.




Note that the expected size of $x$ is approximately $p^2$.