Primitives roots of $p$ and $p^2$. It is well known that for any primitive root $g$ of a prime $p$, either $g$ or $g + p$ is a primitive root of $p^2$
Do there exist any primes for which $g$ is a primitive root of $p^2$ for all primitive roots $g$ of $p$. Does the reverse exist (i.e. $g$ is not a primitive root of $p^2$ for all $g$)? Are there any general conditions for this to be true?
Edit: As Quimey points out, 3 and 2 satisfy the conditions respectively. Are there any other non-trivial examples?
 A: There is a slight problem with the questions, in that if $g$ is a primitive root of $p$, then $g+kp$ should really be considered as being "the same" primitive root of $p$ as $g$.  For $p$ an odd prime, there are exactly $(p-1)\varphi(p-1)$ primitive roots of $p^2$, and $\varphi(p-1)$ primitive roots of $p$. For any primitive root $g$ of $p$, exactly one of the numbers $g+kp$, where $0 \le k \le p-1$, is not a primitive root of $p^2$.
Thus the answer to your question is completely dependent on which numbers we pick as our complete set of representatives of the primitive roots of $p$. So to attack the question, we need to put constraints on this set of representatives. A reasonable constraint that may be the one you intend goes as follows.
Definition: Let $g$ be an integer. We say that $g$ is a small primitive root of $p$ if $1\le g<p$ and $g$ is a primitive root of $p$.
The following little result gives some information about your question:
Let $p$ be a prime of the form $4k+1$. Then there is a small primitive root of $p$ which is also a primitive root of $p^2$.
Proof: Let $g$ be a primitive root of the odd prime $p$. Suppose also that $g$ is not a primitive root of $p^2$. This is the case iff $g^{p-1}\equiv 1\pmod{p^2}$. 
If $p$ is of the form $4k+1$, then $-g$ is also a primitive root of $p$, and $(-g)^{p-1}\equiv 1\pmod{p^2}$. But then $p-g$ is a primitive root of $p^2$.  
Thus for any small primitive root $g$ of a prime of the form $4k+1$, at least one of $g$ or $p-g$ is a primitive root of $p^2$. 
Added: The OP in a comment asked for a proof that if $g$ is a primitive root of $p$, then exactly one of $g+kp$ ($0\le k \le p-1$) fails to be a primitive root of $p^2$. A detailed proof is much too long for a comment, so we added it to the above answer. I have not seen the result proved explicitly, but the proof is almost the same as the usual proof that at least one of $g$ and $g+p$ is a primitive root of $p^2$.
By a counting argument, there are $\varphi(p-1)$ numbers, pairwise incongruent modulo $p^2$, which are primitive roots of $p$ but not of $p^2$. So it will be enough to show that no two of these are congruent modulo $p$.
Suppose to the contrary  $g_1$ and $g_2$ are primitive roots of $p$, but not of $p^2$, are incongruent modulo $p^2$, but congruent modulo $p$.  Then  $g_2=g_1+lp$ for some integer $l$. 
Since the $g_i$ are not primitive roots of $p^2$, but are primitive roots of $p$, they have order $p-1$ modulo $p^2$. Thus $g_1^{p-1}\equiv g_2^{p-1}\equiv 1 \pmod{p^2}$.  By the Binomial Theorem,
$$g_2^{p-1}=(g_1+lp)^{p-1}\equiv g_1^{p-1}+(p-1)(lp)g_1^{p-2}\pmod{p^2}.\tag{$\ast$}$$
But since each of $g_2^{p-1}$ and $g_1^{p-1}$ is congruent to $1$ modulo $p^2$, it follows that $(p-1)(lp)g_1^{p-2}$ is divisible by $p^2$. Thus $l$ is divisible by $p$, and therefore $g_1$ and $g_2$ are congruent modulo $p^2$, contrary to assumption.
We need not have used a counting argument to prove existence from uniqueness. The relationship 
$$g_2^{p-1}\equiv g_1^{p-1}+(p-1)(lp)g_1^{p-2}\pmod{p^2}$$
 can be used directly to show that for any primitive root $g_1$ of $p$, there is an $l$ such that $g_1+lp$ is not a primitive root of $p^2$. It is just a matter of solving the appropriate congruence, like in the proof of Hensel lifting.
