I have some questions about moving from $\mathbb Z_{c^k}$ into $\mathbb Z_{c^{k+1}}$-specifically, with regard to the order of elements.
Suppose $a$ (which is coprime to $c$) has multiplicative order $b$ in $\mathbb Z_{c^k}$.
Now if $a$ has multiplicative order $p$ in $\mathbb Z_{c^{k+1}}$, then, $c^{k+1}\mid (a^p-1)\implies c^k\mid(a^p-1)\implies b\mid p$
So $p=bn$ for some positive integer $n$.
$$\require{cancel}a^b\equiv 1\mod c^k\implies a^{bn}\equiv 1\mod c^{k+1}$$
$$(a^b)^n\equiv 1\mod c^{k+1}\implies ((a^b-1)+1)^n\equiv 1 \mod c^{k+1}$$
$$\sum_{i=0}^n\Biggl(\binom{n}{i}(a^b-1)^i1^{n-i}\Biggr)\equiv 1 \mod c^{k+1}$$
Since $c^k|(a^b-1):$
$$\binom{n}{0}(a^b-1)^01^{n-0}+\binom{n}{1}(a^b-1)^11^{n-1}+\sum_{i=2}^n\cancelto{0}{\Biggl(\binom{n}{i}(a^b-1)^i1^{n-i}\Biggr)}\equiv 1\mod c^{k+1}$$
$$1+n(a^b-1)\equiv 1\mod c^{k+1}$$
So, finally:
$$n(a^b-1)\equiv 0\mod\ c^{k+1}$$
This gives us that $n|c$. More specifically, if $g.c.d.\Bigl(\frac{a^b-1}{c^k},c\Bigr)=d$, then $n=\frac{c}{d}$.
Now, looking at something like the base-$a$ Wieferich primes, there are clearly examples when $n\neq c$ (technically, Wieferich primes don't care specifically about the multiplicative order of the bases, but it happens frequently enough that $p-1$ is the base's order).
I have two main questions. For any non-perfect power, $c$, can we always find an $a$ and a $k$ such that, if $ \operatorname{ord}(a)=b$ in $\mathbb Z_c$ and $ \operatorname{ord}(a)=n$ in $Z_{c^k}$, $n\neq bc^{k-1}$? And, second, (with $c$ still being a non-perfect power$\neq2$) is it possible that, if, for some given $a$, that $ \operatorname{ord}(a)=b$ in $\mathbb Z_c$ and $ \operatorname{ord}(a)=bc$ in $\mathbb Z_{c^2}$, but for some $k>2$, $ \operatorname{ord}(a)\neq bc^{k-1}$ in $\mathbb Z_{c^k}$?
Edit: I think I have an answer to the first question-it looks to be a bit more general.
If we look at the set of elements in $\mathbb Z_{c^{k+1}}$ that are invertible, we have a group under the operation of multiplication- denote it by $V_{c^{k+1}}$. Of course, these are exactly the elements that are coprime to $c^{k+1}$ so the order of this group is $$\phi(c^{k+1})=c^{k+1}\prod_{p:p=prime,p\mid c^{k+1}}\Bigl(1-\frac{1}{p}\Bigr)=c^{k+1}\prod_{p:p=prime,p\mid c}\Bigl(1-\frac{1}{p}\Bigr)=\\\frac{c^{k+1}}{\prod_{p:p=prime,p\mid c}p}\prod_{p:p=prime,p\mid c}(p-1)=d\prod_{p:p=prime,p\mid c}(p-1)$$
where $d$ is clearly an integer. Now, consider any prime factor, $q$, of $p_{min}-1$ (where $p_{min}$ is the smallest prime that divides $c$).
$$q\mid (p_{min}-1)\mid\biggl(\prod_{p:p=prime,p\mid c}(p-1)\biggr)\mid \phi(c^{k+1})=\Bigl|V_{c^{k+1}}\Bigr|$$
So $q\mid \Bigl|V_{c^{k+1}}\Bigr|$. But by Cauchy's theorem for groups, this means that $V_{c^{k+1}}$ has an element of order $q$- denote it by $x$. Clearly the multiplicative of order of $x$ in $\mathbb Z_{c^{k+1}}$ is also $q$. Referring back to the way $q$ was defined, it's clear that $q$ and $c$ are coprime (since $q$ must be less than any prime factor of $c$). But, if $x$ has order $b$ in $\mathbb Z_{c^k}$, then the order of $x$ in $\mathbb Z_{c^{k+1}}$ is (as shown before) of the form $bn$ where $n$ is some divisor of $c$.
That means $bn=q$. Since $q$ and $c$ are coprime, $n=1$, and $b=q$, i.e., the order of $x$ is the same in $\mathbb Z_{c^{k+1}}$ as it is in $\mathbb Z_{c^k}$.
The reason this seems more general is that it works for any $k$. Of course, there's a small flaw when it comes to $c$ being even, i.e. that $p_{min}-1=2-1=1$ always. But, the above argument could still easily be extended to any even numbers where the $2^{nd}$ smallest prime divisor isn't a Fermat prime (just let $q$ be some prime other than $2$ that divides the second smallest prime-1). Even if the second smallest prime divisor is Fermat, we'll still have $\operatorname{ord}(a)=2b\neq bc^k$ (I just can't guarantee an element with the same order in $\mathbb Z_{c^k}$ and $\mathbb Z_{c^{k+1}}$ exists). That seems to just leave $c=2$. Still have no clue how to deal with the second question.
Edit 2: The above argument (in the Edit) can be tweaked a bit- once we have an $x$ of order $q$ in $\mathbb Z_{c^{k+1}}$, instead of returning to $\mathbb Z_{c^k}$ and examining $x$'s order there, we can look directly at $\mathbb Z_c$ instead. If the order of $x$ in $\mathbb Z_c$ is $b$, then by extending the reasoning presented above, the order of $x$ in $\mathbb Z_{c^{k+1}}$ is $bN$, where $N$ is a divisor of $c^k$. That means $bN=q$ and (since $q$ and $c$ are coprime) $b=q$ but now $b$ is the order of $x$ in $\mathbb Z_c$. This reasoning can be extended so that for every $1\leq h\leq k+1$, the order of $x$ in $\mathbb Z_{c^h}$ is the same. This seems to be suggesting that the answer to the second question is generally no- but that would be to assume that elements that do not have order $bc^k$ in $\mathbb Z_{c^{k+1}}$ are only constructed by the method in the first Edit (it also isn't true in the case when $c=2$ just by the fact that all odd numbers have order 1 in $\mathbb Z_2$).