TITLE: Can we find all primes $p$ that we can find a number $a \neq 1$ that has the same order in $\mathbb{Z}_p^{*}$ and $\mathbb{Z}_{p^2}^{*}$
QUESTION [5 upvotes]: Supposed that there is a prime $p$, we may define $\mathbb{Z}_p^{*}$ as the multiplication group of integer modulo $p$. Each element in $\mathbb{Z}_p^{*}$, $\{1, 2, \ldots, p-1\}$, has finite order. Meanwhile, if we consider $\mathbb{Z}_{p^2}^{*}$, all $a \in \{1, 2,\ldots, p-1\}$ also have finite order in $\mathbb{Z}_{p^2}^{*}$, but the order of $a$ could be different from the order of respective $a$ in $\mathbb{Z}_p^{*}$.
The problem is that, could we find all prime $p$ that there is a number $a \in \{2, \ldots, p-1\}$ that the order of $a$ in $\mathbb{Z}_p^{*}$ is the same as the order of $a$ in $\mathbb{Z}_{p^2}^{*}$?
Edit: $a = 1$ definitely satisfies the condition for all primes $p$, so we should consider if there is another nontrivial one which satisfies the property.

REPLY [2 votes]: The primes you want are listed at OEIS/A134307. The first 20 ones are
$$
11, 29, 37, 43, 59, 71, 79, 97, 103, 109, 113, 127, 131, 137, 151, 163, 181, 191, 197, 199 
$$
Not much is known about them. Heuristically, the density of such primes is $1 - \dfrac1e \approx 0.632$.