For this question, suppose $p = 659$, where $p$ is prime. I have found that, through computing $\phi(p-1)$, where $\phi ()$ is the Euler Tiotent Function. Here, there are a total of 276 primitive roots modulo $659$. Here is my question:
How many elements are there of order $94$ modulo $p?$ How about order $49?$
I will focus on the former. I know that the order of modulo $p$ is the defined as the quantity
$e_{p}(a) =$ (the smallest exponent $e \geq 1$ such that $a^e \equiv 1$ (mod $p$). So I need to find which $e's$ satisfy this condition such that $94^e \equiv 1$ (mod $659$). This is what I believe is the discrete logarithm problem, which makes it difficult to solve for $e$. How else can I go about solving this problem?
So (with $p$ a prime) the non-zero elements modulo $p$ form a cyclic group of order $p-1$ generated by any of the $\varphi(p-1)$ primitive roots modulo $p$.
A finite cyclic group has exactly one subgroup of each order $q$, where $q$ is a factor of $p$ and $p=qr$. If $a$ is a primitive root, this subgroup is generated by $a^r$.
In your example all the elements of order $94$ belong to the same cyclic subgroup of order $94$, and this has $\varphi (94)$ generators of order exactly $94$. You can reach this conclusion without ever displaying an element of order $94$ (so no need to solve the discrete logarithm problem, or anything like that). If you know a primitive root, you can find such an element quite easily ($a^r$ - see above).
In your question you have drifted into looking for the order of $94$ rather than the number of elements of that order. You need to keep your mind on the problem in hand. The discrete logarithm problem says given a primitive root $a$, how can I find an $s$ with $a^s= \text{ say } 94$, but this is not the same as finding an element of order $94$.