Haar Measures and Embeddings of $\nu$-adic integers

Let $$\nu\geq4$$ be any composite integer, and let $$d\in\left\{ 2,\ldots,\nu-1\right\}$$ be any non-trivial divisor of $$\nu$$. Since $$d\mid\nu$$, note that any sequence $$\left\{ \mathfrak{y}_{n}\right\} _{n\geq1}\subseteq\mathbb{Z}_{\nu}$$ which converges in $$\nu$$-adic absolute value to a limit $$\mathfrak{y}$$ necessarily converges to $$\mathfrak{y}$$ with respect to the $$d$$-adic absolute value, as well. Since $$\mathbb{Z}_{\nu}$$ (resp. $$\mathbb{Z}_{d}$$) is simply the closure/completion of $$\mathbb{Z}$$ with respect to the $$\nu$$-adic (resp. $$d$$-adic) topology, this implies that the inclusion $$\iota:\mathbb{Z}_{\nu}\hookrightarrow\mathbb{Z}_{d}$$ is an embedding of $$\mathbb{Z}_{\nu}$$ in $$\mathbb{Z}_{d}$$.

Now, let $$\mu_{\nu}$$ (resp. $$\mu_{d}$$) be the Haar measure on $$\mathbb{Z}_{\nu}$$ (resp. $$\mathbb{Z}_{d}$$), normalized so that $$\mu_{\nu}\left(\mathbb{Z}_{\nu}\right)=1$$ (resp. $$\mu_{d}\left(\mathbb{Z}_{d}\right)=1$$). How does the inclusion map $$\iota$$ affect the measures of sets? That is, given a measurable set $$V\subseteq\mathbb{Z}_{\nu}$$, and letting $$\iota\left(V\right)$$ denote the copy of $$V$$ embedded in $$\mathbb{Z}_{d}$$, how does $$\mu_{\nu}\left(V\right)$$ compare to $$\mu_{d}\left(\iota\left(V\right)\right)$$?

My intuition is that $$V$$ should be $$\mu_{d}$$-measurable, and that $$\mu_{d}\left(\iota\left(V\right)\right)$$ should be $$\geq\mu_{\nu}\left(V\right)$$ (i.e., $$\mathbb{Z}_{\nu}$$ is a “small” subset of $$\mathbb{Z}_{d}$$). I'm wondering if this is correct, and if it can be taken further; for instance, is $$\mu_{d}\left(\iota\left(V\right)\right)=0$$ for all $$V\subseteq\mathbb{Z}_{\nu}$$? An answer and/or a reference to an answer would be much appreciated.

The contrapositive statement of Steinhaus' Theorem says that for a locally compact abelian group $$G$$ (written additively), a haar-measurable subset $$A\subseteq G$$ will have zero measure in $$G$$ if $$A-A=\left\{ a-b:a,b\in A\right\}$$ does not contain an open neighborhood of the identity element of $$G$$.
So, letting $$G=\mathbb{Z}_{d}$$ and letting $$A=\mathbb{Z}_{\nu}$$, define for each positive integer $$N$$ the $$d$$-adic number:
$$\mathfrak{c}_{N}=\sum_{n=N}^{\infty}d^{n}\in\mathbb{Z}_{d}$$
Note that $$\mathfrak{c}_{N}\rightarrow0$$ $$d$$-adically as $$N\rightarrow\infty$$. If $$\nu$$ is not of the form $$d^{m}$$ for some integer $$m\geq1$$, it follows that the $$\mathfrak{c}_{N}$$s are a sequence of elements in $$\mathbb{Z}_{d}\backslash\mathbb{Z}_{\nu}$$ which converge to $$0$$ in $$\mathbb{Z}_{d}$$. Since $$0$$ is therefore an accumulation point of the complement of $$\mathbb{Z}_{\nu}$$ in $$\mathbb{Z}_{d}$$, the set $$\mathbb{Z}_{\nu}-\mathbb{Z}_{\nu}=\left\{ \mathfrak{a}-\mathfrak{b}:\mathfrak{a},\mathfrak{b}\in\mathbb{Z}_{\nu}\right\}$$ does not contain a $$d$$-adically open neighborhood of $$0$$. Thus, by Steinhaus' Theorem, $$\mathbb{Z}_{\nu}$$ must have zero measure in $$\mathbb{Z}_{d}$$ whenever $$\nu$$ is a positive integer multiple of $$d$$ which is not a power of $$d$$.