Changing uniformizer of $p$-adics In the theory of $p$-adic fields typically a uniformizer $\pi$ is chosen that generates the maximal ideal, $m$. And a few theorems later it can be shown that every element $x \in O$ of the ring of integers can be expressed as $$x = \sum_{i=0}^\infty c_i \pi^i$$ where the $c_i$ are representatives of $O/m$. But $\pi$ is only unique up to multiplication by a unit.

Question 1: What is the proof that the $p$-adic field does not depend on the choice of uniformizer? (Edit: This was shown in the comments to be a silly question, but I believe the following questions are still valid.)
Question 2: Suppose $\pi_2=u\pi$ for some unit $u$. Is there an explicit form of
   the mapping $c_i \to d_i$ such that $$\sum_{i=0}^\infty c_i \pi^i =
 \sum_{i=0}^\infty d_i \pi_2^i$$

As an example, consider $\mathbb{Z}_7$, the 7-adics. The usual uniformizer is, of course, $\pi=7$. But $5$ is a unit in $\mathbb{Z}_7$ so we should be able to take $\pi=35$ or $\pi=7/5$.

Question 2 (example version): Is there an explicit form of the mapping $c_i \to d_i$ such that $$\sum_{i=0}^\infty c_i 7^i = \sum_{i=0}^\infty d_i (7/5)^i$$ Note: The answer is not $d_i = 5^i c_i$ because $0 \le d_i \le 6$.

 A: As regards question 2, this happens to be exactly what I tackled in my answer to Why does this generalized ring of Witt vectors not depend on a choice of a prime element? There, in an intro, I tried to write a $5$-adic integer in base $10 = 2\cdot 5$ (as $2$ is a $5$-adic unit). To quote the crucial part of the general statement:

To translate from "standard base $p$" to "base $\tilde p = u\cdot p$" with a unit $u \in \mathbb Z_p^\times$, looking through those congruences one has to solve, I get (denoting $[\cdot] : (\mathbb Z_p \rightarrow) \mathbb Z_p/p \rightarrow \{0,1,\dots , p-1\}$ our representatives):
$\tilde a_0 = a_0$ ($= [a_0 \text{ mod } p]$)
$\tilde a_1 = \left[ [u^{-1}] [a_1] \qquad \text{ mod } p \right]$
$\tilde a_2 = \left[ [u^{-2}] \left([a_2]+ \dfrac{1-[u^{-1}]u}{p} \cdot [a_1] \right) \qquad \text{ mod } p \right]$
and it gets unwieldy from here, but if so obliged, one will find formulae for the higher terms. Note that although we start and end with expressions modulo $p$, we have to go through computations in $\mathbb Z_p$ here.
In other words, we can define a map $\mathbb F_p^{\mathbb N_0} \rightarrow \mathbb F_p^{\mathbb N_0}$,  $(a_0 \text{ mod }  p, a_1 \text{ mod } p, ...) \mapsto (\tilde a_0 \text{ mod } p, \tilde a_1 \text{ mod }p, ...)$, such that the induced map on $\mathbb Z_p (\simeq W(\mathbb F_p))$,
$$\sum a_n p^n \mapsto \sum \tilde a_n \tilde p^n$$
is a ring homomorphism. Actually, it is the identity, written in an unpleasantly complicated way.

I believe that is what you are looking for.
I want to note that the "right" setup for this seems to be the Witt vector machinery in the follow-up in that answer, which gives general polynomials $T_0, T_1, ...$ whose manifestations are the above formulas (and you want to apply it to $R = O/m$ which should give $W(R) \simeq O$ regardless of whether $W(\cdot)$ is defined with $\pi$ or $\tilde \pi = u \cdot \pi$). One thing that makes the above formulas uglier are those representative choices $[ \cdot ]$. Note that for all this, it is tacitly supposed that in both extensions, in your terminology $\sum c_i \pi^i$ and $\sum d_i \pi^i$, the $c_i$ and $d_i$ come from the same set of representatives, chosen as one section $[ \cdot ] : O/m \rightarrow O$ of the projection $O \rightarrow O/m$; and conceptually, this choice of representatives brings an unwarranted arbitrariness into the question. (The Witt vector formulation, on the other hand, once one has those general polyomials $T_n$, seemingly just works in the residue field.)
