$(\mathbb{E}^n,+)$ is embedded in the Group $(M(n),\circ)$ as its Normal subgroup.
Define $T:\mathbb{E}^n \to M(n)$ be the map $T(v):=T_v$, the Translation defined by the point $v$. i.e. $T_v(x)=x+v$ for all $x \in \mathbb{E}^n.$
Claim 1: $T$ is a group homomorphism between $\mathbb{E}^n$ and the group $M(n)$
Let $u,v \in \mathbb{E}^n.$ then $T(u+v)=T_{u+v}, T(u)=T_u$ and $T(v)=T_v$. Now for any $x \in \mathbb{E}^n$ we have the following: $$T_{u+v}(x) = x+(u+v) = x+(v+u) = (x+v)+u = T_{v}(x)+u = T_u(T_v(x)) = T_u \circ T_v(x)$$ $$\implies T(u+v) = T(u) \circ T(v) \ \text{for all} \ u,v \in \mathbb{E}^n $$ Therefore $T$ is a Group homomorphism.
Claim 2: $T$ is One-One and $T(\mathbb{E}^n)$ is a Normal subgroup of $M(n)$.
For $u,v \in \mathbb{E}^n$ we have $$T_u = T_v \Rightarrow T_u(x) = T_v(x) \ \text{for all} \ x \in \mathbb{E}^n \Rightarrow T_u(0) = T_v(0) \Rightarrow u=v.$$ Hence $T$ is One-One.
Note that $T(\mathbb{E}^n)=\{T_u:u \in \mathbb{E}^n\}$ i.e all Translations of $\mathbb{E}^n$. Let $S \in M(n)$ and choose any $T_u$ from the image set, then I have to show that $S\circ T_u \circ S^{-1}$ is a Translation. i.e.
$$S\circ T_u \circ S^{-1} (x) = x+v = T_v(x) \ \text{for some} \ v \in \mathbb{E}^n.$$
How can I show that?
Although there are a few answers here, but if someone helps me solve the way I proceeded above. It will be very helpful. Thank you.
An isometry of euclidean space is an affine transformation $S$ of the form $S(x) = Ox + p$ where $O$ is an orthogonal matrix. The inverse can be computed explicitly (in terms of $O$ and $p$) as
$$ S^{-1}(x) = O^{-1}x - O^{-1}p. $$
With this in mind, you just have to compute the conjugation of $T_u$ by $S$,
$$ ST_u S^{-1}(x) = ST_u(O^{-1}x - O^{-1}p) = S(O^{-1}x - O^{-1}p + u) = O(O^{-1}x - O^{-1}p + u) + p = T_{Ou}(x). $$
Edit: an isometry $f$ is an affine map because of the following: take $x,y$ in Euclidean space and let $d = \|x-y\|$. By preservation of distances, for each $t \in [0,1]$ we have
$$ f(tx+(1-t)y) = f(\overline{B}(x,td) \cap \overline{B}(y,(1-t)d)) = \overline{B}(f(x),td) \cap \overline{B}(f(y),(1-t)d) = tf(x) + (1-t)f(y). $$
Now, the map $fT_{-f(0)}$ is still an isometry and it is linear. A standar argument in linear algebra shows that it is represented by an orthogonal matrix, thus $f(x) = Ox+p$ where $O \in O(n)$ and $p=f(0)$.