Is the image in $\mathbb{Q}_{p}^{\times}$ of a Cauchy sequence in $\mathbb{Q}_{p}$ in general, also Cauchy? Must it be a fixed point?
I'm assuming that it's conventional to consider a sequence to be Cauchy in $\mathbb{Q}_p$ if it's Cauchy as measured by its p-adic metric.
By the image in $\mathbb{Q}_p^{\times}$ of a number $n$ in $\mathbb{Q}_p$ I mean $n\times\lvert n\rvert_p$ for example the image of $24$ in $\mathbb{Q}_2^{\times}$ is $3$.
If true, this fact would be helpful. It looks like it should generally be true but if we consider the sequence:
$2^n\times n\times (2(n\mod 2)+1)$ for increasing integers $n$, i.e. the sequence:
$\{1,6,4,24,16,96,64,\ldots\}$
We can see from the sequence $\lvert n_i-n_{i-1}\rvert_2=\{1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},\ldots\}$ that it is Cauchy in $\mathbb{Q}_2$ despite the fact that its image in $\mathbb{Q}_2^{\times}$ is $\{1,3,1,3,1,\ldots\}$ and it would appear on the face of it to have two distinct trajectories.
Now I'm led to believe that the 2-adics are something of a special case when it comes to convergence because uniquely their logarithm has a nontrivial kernel in the $\mathbb{Z}_2$-principal units (and let me be clear, I have next to no idea what that means). The question is, is this independence of convergence in the two components of Cauchy Sequences unique to $\mathbb{Q}_2$ or can this happen for any $p$?
No, there's no such requirement in any base $p$. Convergence of a sequence in $\mathbb{Q}_p$ and convergence of its image in $\mathbb{Q}_p^{\times}$ can be independent, as demonstrated by:
The Cauchy sequence $(-p)^k$ whose image in $\mathbb{Q}_p^{\times}:\{-1,+1\}$ does not converge.
And the sequence:
$\{2,12,14,60,62\ldots\}$ which does not converge in $\mathbb{Q}_2$ while its image in $\mathbb{Q}_2^{\times}: \{1,3,7,15,31,\ldots\}$ is Cauchy.