Here is how I think of it:
A METRIC is a "spatial thing", it tells us "how separated" (we use "how far apart" as measured by the metric) two points are.
A NORM is a "vector thing"-there is this requirement of "homogeneity" (also called absolute scalability)which tells us that the norm is "consistent" with scalar multiplication:
$\|\alpha v\| = |\alpha|\|v\|$
Given a normed vector space, one can define a metric $d$ by:
$d(u,v) = \|u - v\|$.
Now one of the properties of a norm $\|\cdot\|$ is that:
$\|v\| = 0 \iff v = 0$.
This implies $d(u,v) = 0 \iff u = v$.
A norm must also satisfy the TRIANGLE INEQUALITY:
$\|u + v\| \leq \|u\| + \|v\|$
From this we have:
$0 = \|v + -v\| \leq \|v\| + \|-v\| = \|v\| + \|(-1)v\| = \|v\| + |-1|\|v\| = 2\|v\|$
so a norm is positive-definite, and so $d: V \times V \to \Bbb R_0^+$.
The triangle property of a norm tells us:
$d(u,w) = \|u - w\| = \|u - v + v - w\| \leq \|u - v\| + \|v - w\| = d(u,v) + d(v,w)$, so we have a bona-fide metric.
However, not ALL metrics on a vector space are "compatible" with the vector space structure:
We can define, for example, the "discrete metric":
$d(v,v) = 0$, for all $v \in V$.
$d(u,v) = 1$, if $u \neq v$, which is not absolutely scalable, so does not correspond to a norm.
It is possible to give examples of two metrics which induce the same topology on a vector space $V$, but one is a norm, and one is not, but I will not do so here.
Even MORE restrictive, is the case of an inner product space. An inner product $\langle \cdot,\cdot\rangle$ induces a norm by:
$\|v\| = \sqrt{\langle v,v \rangle}$
But not all norms arise in this way, an example is the $p$-norm for $p \neq 2$:
$\displaystyle \|v\|_p = \left(\sum_i^n |v_i|^p\right)^{1/p}$ which is NOT an inner product.
$\Bbb R^n$ is rather special: It's an inner product space with a topology induced by the metric which is induced by the norm induced by its inner product. Moreover, the vector space addition and scalar multiplication are continuous maps.
Note that $\text{End}(\Bbb R^n)$ has a natural ring structure (the ring of Endomorphisms of the $\Bbb R$-module $\Bbb R^n$), which is isomorphic to the ring of $n \times n$ matrices over $\Bbb R$ (this isomorphism is not "canonical" but depends on a choice of basis, so speaking in terms of "matrices" depends on picking a "coordinate system", while speaking of $\text{End}(\Bbb R^n)$ is "basis-free")).
The group of units of this ring, is the general linear group, which can also be written $\text{Aut}(\Bbb R^n)$. Note that LINEAR isometries (under any metric for for a finite-dimensional vector space) are automatically bijective (isometries are always injective, via positivity of the metric, and in a finite-dimensional vector space setting, $T \in \text{End}(V)$ is injective if and only if it is surjective, by the rank-nullity theorem).
In the "physical world", invertible linear maps correspond to "reversible" linear transformations of a space: in other words, "no loss of information". In more abstract structures, just as with ordinary integers, 0 continues to play "the bad guy".