Is there any relation between \(\displaystyle\text{Sym}_{{n}}\) and

Step 1
${\displaystyle S_n}$ is not contained in ${\displaystyle S_m}$ when ${\displaystyle n<m}$ if you define ${\displaystyle S_k}$ to be the set of all bijections ${\displaystyle \{1,\cdots ,k\}\to \{1,\cdots ,k\}}$, simply because an element of ${\displaystyle S_n}$ has domain ${\displaystyle \{1,\cdots ,n\}}$ and elements of ${\displaystyle S_m}$ have domain ${\displaystyle \{1,\cdots ,m\}}$, and the two sets are different.
Now this is a silly problem, so surely there is some way out.
Suppose again that ${\displaystyle n<m}$. There is map ${\displaystyle \varphi :S_n\to S_m}$ such that whenever ${\displaystyle \pi \in S_n}$, so that ${\displaystyle \pi :\{1,\cdots ,n\}\to \{1,\cdots ,n\}}$ is a bijection, then ${\displaystyle \varphi \left(\pi \right):\{1,\cdots ,m\}\to \{1,\cdots ,m\}}$ is the map given by
$\varphi (\pi )(i)=\{\begin{array}{ll}\pi (i),& \text{if }i\le n\text{;}\\ i,& \text{if }i>n\text{.}\end{array}$
You can easily check that ${\displaystyle \varphi }$ is well-defined (that is, that ${\displaystyle \varphi \left(\pi \right)}$ is a bijection for all ${\displaystyle \pi \in S_n}$) and that moreover ${\displaystyle \varphi }$ is a group homomorphism which is injective.
It follows from this that the image of ${\displaystyle \varphi }$ is a subgroup of ${\displaystyle S_m}$ which is isomorphic to ${\displaystyle S_n}$. It is usual to identify ${\displaystyle S_n}$ with its image ${\displaystyle \varphi \left(S_n\right)\subseteq S_m}$, and then we can say that «${\displaystyle S_n}$ is a subgroup of ${\displaystyle S_m}$», but this is nothing but a façon de parler.
It should be noted, though, that there are many injective homomorphisms ${\displaystyle S_n\to S_m}$, so there are many ways to identify a subgroup of ${\displaystyle S_m}$ with ${\displaystyle S_n}$. The one I described above is nice because it looks very natural.

Step 1
${\displaystyle Sy{m}_k<Sy{m}_n}$ for ${\displaystyle n<k}$ and it corresponds to a subgroup of all permutations with the same ${\displaystyle k-n}$ elements.
(That is, fix ${\displaystyle k-n}$ elements first, then look at all the transpositions that keep them in place. This subgroup of ${\displaystyle S_k}$ is isomorphic to ${\displaystyle S_n}$.)