Motivation: In $\langle\mathbb{Z}/7\mathbb{Z},\times\rangle,\ \langle 3\rangle = (3,2,6,4,5,1).$
Given a $k-$tuple of distinct integers, $q_1, q_2, \ldots, q_k,$ (all nonzero) does $\exists$ integers $a,m,n$ with $n>\max\{ q_1,q_2,\ldots,q_k\},\ $ such that $a m^i \equiv q_{i} \pmod n\ \forall\ i\in \{1,2,\ldots, k\}\ $ [In other words, $q_i$ is the remainder when $am^i$ is divided by $n.]\ ?$
Maybe there is a counter-example like $(100,99,98, 3,2,1).$ But I doubt it. I imagine it is true and we can even take $a=1,$ for example. I think it could be true by Chinese remainder theorem and/or Fermat's Little theorem?
It should be true for $k=2,$ so maybe then we can use induction on $k$?
I think that without the "nonzero" and "all distinct" conditions, there would be easy counter-examples.
Edit: I think $n$ must be prime for this to work, right? Well, I think that if $n$ is prime then every number other than $0$ and $1$ "generates" all of $\{1,2,\ldots,n-1\},$ although I don't recall the proof for this. But is it true that if n is not prime, then in $\langle\mathbb{Z}/n\mathbb{Z},\times\rangle,\ $ then for any $u,\ \langle u\rangle\ $ does not generate all of $\{1,2,\ldots,n-1\}\ ?$ Can someone link to proofs of these two theorems? I'm sure they are in some Elementary Number/Group Theory pdf. I also have the book, "Elementary Number Theory" by Springer, but I'm not sure if these theorems/proof are in there...
Basically, you want geometric sequence modulo some $n$, so $q_i^2 \equiv q_{i-1}q_{i+1}\pmod n$, i.e. $n\mid q_i^2-q_{i-1}q_{i+1}$, so to find counterexample you can take any triple $(q_1,q_2,q_3)$ such that $q_2^2-q_1q_3 = 1$, for example $q_1 = 1, q_3 = q_2^2-1$ with $q_2$ arbitrary.
In fact, take any three consecutive integers $x-1,x,x+1$, then $x^2\equiv (x-1)(x+1) \pmod n$ iff $n\mid 1$, so we have plenty of counterexamples.
For $k = 2$, the answer is almost trivial, take any prime $n$ greater tghan $q_1$ and $q_2$ and let $m \equiv q_2q_1^{-1}$, $a\equiv q_1m^{-1}$, where the inverse is modulo $n$. Or, slightly amusingly, take $n = q_1+q_2$, $m=-1$ and $a = q_2$. Then $q_1\equiv am\pmod n$, $q_2\equiv am^2\pmod n$.
For the questions in edit, this depends on what you mean by generate. If by generate you mean you take the smallest ideal containing generator, then for a prime $p$, ring $\mathbb Z/p\mathbb Z$ is a field by Bézout's identity, since if $a\not\equiv 0 \pmod p$, $a$ and $p$ are relatively prime, so there are integers $x,y$ such that $ax+py = 1$ and therefore $ax \equiv 1 \pmod p$, so $x \equiv a^{-1} \pmod p$. The only ideals of a field are $(0)$ or the field itself, so all non-zero elements in $\mathbb Z/p\mathbb Z$ generate the whole field.
If by generate you mean take the smallest multiplicative subgroup of $(\mathbb Z/p\mathbb Z)^\times$ containing generator, for non-zero $g$ in $\mathbb Z/p\mathbb Z$ you want $\{g,g^2,\ldots,g^{p-1}\}$ to be all distinct, but this happens precisely when the order of $g$ in $(\mathbb Z/p\mathbb Z)^\times$ is $p-1$. This doesn't need to be the case for all non-trivial $g$, for example $4^2\equiv 1 \pmod 5$, so $\bar 4$ has order $2$ in $(\mathbb Z/5\mathbb Z)^\times$ and in general $\overline{p-1}$ has order $2$ in $(\mathbb Z/p\mathbb Z)^\times$ since $(p-1)^2 \equiv p^2-2p+1 \equiv 1 \pmod p$. Taking some non-zero $a$ and considering $\{ag,ag^2,\ldots, ag^{p-1}\}$ won't change anything since multiplication by $a$ is bijection on $\mathbb Z/p\mathbb Z$.
Of course, when $n$ is not prime, take a non-trivial divisor of $n$.