How hard is Self-DLOG? This question while asking for something different brings up an intriguing problem: If you can find $x$ such that $x\equiv R^x\pmod p$ then you can break DSA. Now I thought that one might be able to relate this to the discrete logarithm problem somehow, but I failed to find the relation and so I'm asking this (somewhat more general) question.

Now my question is:
Let $p$ be an arbitrary prime, does there exist an algorithm faster than $\mathcal O(p)$ which, given $p$ and an arbitrary $R\in\mathbb Z$, finds an $x\in\{y\in\mathbb Z\mid 0\leq y&lt;p\}$ such that $x\equiv R^x\pmod p$ if such an $x$ exists or $\perp$ otherwise?
The problem about fixed  points of discrete logarithms is  known the Brizolis problem. In particular for the average number of solutions
$$N(p)=\frac{1}{\varphi(p-1)}\sum_g\left|\{0\le x\le p-1:g^x=x\pmod p\}\right|$$ is known that (Grechnikov, 2012) $N(p)=1+S(p)$, where 
$$-C(\varepsilon)p^{-1/4+\varepsilon}\le S(p)\le \exp(C'\mathrm{Li}((\log p)^{c\frac{\log\log\log\log p}{\log\log\log p}})).$$

It is also necessary to say that inversed problem is easier: for a given $x$ one can take $R=x^{x^{-1}\bmod (p-1)}.$