Any way to find $g,P$ for max cycle size in Blum–Micali with $x_{i+1} = g^{x_i} \mod P $ and $x_0 = g$?

For some $g$ and prime $P$ the sequence $$x_{i+1} = g^{x_i} \mod P $$ $$ x_0 = g$$ can contain all numbers from $1$ to $P-1$ and with this it is a pseudo-random permutation of those numbers (EDIT: seems to be not the case).

Is there any (quick) way to find big/safe values for $P$ and related $g$ which can still produce every number from $1$ to $P-1$?

Some examples:

With $P=5, g=3$ the sequence would be $$\begin{split} &[3, 3^3\equiv 2, 3^{2} \equiv 4, 3^{4} \equiv 1] \mod 5 \\ \equiv&[3, 2, 4, 1] \mod 5 \end{split}$$

Or for $P=23, g=20$ the values would be: $$[20,18,2,9,5,10,8,6,16,13,14,4,12,3,19,17,7,21,15,11,22,1]$$ or $P=59, g=39$

It looks like not every $P$ has such a value $g$. In some test run small $P$ often had no more than one suitable $g$.
[ 107: 94]
[ 359: 97]
[ 467: 72]
[ 587: 150,375]
[ 719: 284]

So far only $P=587$ got more than one $g$ in my test run. (Edit: I only checked for $P=2q+1$ with $q$ a prime, other $P$ can work as well)

side questions:
Will multiple $g$ be more common for larger $P$?
Or will larger $P$ tend to have no $g$ at all?

main question:
Is there any (quick) way to find big/safe values for $P$ and a related $g$ ?

I'm afraid that I can offer little the way of proofs, but I do have some observations and heuristics.

Firstly the map will only be a permutation if $g$ is a primitive root modulo $P$. We note that there are $\phi(P-1)$ primitive roots and that primes with many primitive roots will have more $g$ for which the permutation might be a full cycle. The primes with the greatest proportion of primitive roots are of the form $P=2q+1$ where $q$ is also prime. Note that all of your examples are of this form.

Next we note that not all permutations are possible as we will always have the sequence $P-1\mapsto 1\mapsto g$. Only a proportion $1/(P-1)(P-2)$ of permutations will have this property and only $1/(P-2)(P-3)$ full cycles will have this property. We also note that even values always map to quadratic residues and odd values always map to quadratic non-residues (with further restrictions for other multiples/powers that divide $P-1$). This is a more powerful restriction which only a proportion $1/\binom{P-1}{(P-1)/2}$ of permutations having this property. It's not immediately clear to me what proportion of full cycles will meet his restriction.

IN PROGRESS