Isomorphisms that have names (Or: Base-less isomorphisms) The answer to this question of mine provided me with the fact that every isomorphism  $$
 \phi: K^{n} \rightarrow V,
$$
where $V$ is an arbitrary vector space of finite dimension $n$ over
the field $K$, has  the form 
$$\begin{eqnarray*} & \phi\left(x_{1},\ldots,x_{n}\right)= x_{1}\vec{v}_{1}+\ldots+x_{n}\vec{v}_{n},
\end{eqnarray*}$$
where $\left(\vec{v}_{1},\ldots,\vec{v}_{n}\right)$ is  some basis
of $V$.
Now I was wondering, is it possible to exhibit an isomorphism from which one can't immediatly guess the basis  $\left(\vec{v}_{1},\ldots,\vec{v}_{n}\right)$  in which its written in ?
$$ $$
Consider this analogy: Every linear mapping $f$ from, say $\mathbb{R}^2$ to $\mathbb{R}^2$ , has the form
 $$ f(x_1,x_2)=(t_1x_1+t_2x_2)\vec{e}_{1} +(t_3x_1+t_4x_2)\vec{e}_{2},$$
where $t_1,t_2,t_3,t_4$ are some real numbers.
A rotation of $45^{\circ}$ counter-clockwise direction is a linear mapping, but it is not obvious , what the $t_1,t_2,t_3,t_4$  should be. (Only after some tinkering one gets that they are $\cos \frac{\pi}{2},-\sin \frac{\pi}{2},\sin \frac{\pi}{2},\cos \frac{\pi}{2}$)
Nonetheless for the typical linear mappings like
$$ f(x_1,x_2)= (x_1,x_2)$$ we won't have such a "form-less" interpretation like saying "it's a rotation", since here it is obvious, that $t_1=1,t_2=0,t_3=0,t_4=1$.
(A different example of a linear mapping for which the $t_1,t_2,t_3,t_4$ aren't obvious to guess is the differentiation operator, that maps every polynomial of degree at most $1$ to its derivative)
 A: No guess is required, you can compute the basis from $v_i = \phi(e_i)$. For example, if $f : \mathbb{R}^2 \to \mathbb{R}^2$ is the rotation you mentioned, we have (by elementary trigonometry) $f(e_1) = (\sqrt{2}/2,\sqrt{2}/2)$ and $f(e_2) = (-\sqrt{2}/2,\sqrt{2}/2)$. Your second example $f=\mathrm{id}$ is the rotation by 0°.
EDIT: I think now I've understood your question. Well, there are tons of examples and occur everywhere in algebraic-tasty branches of mathematics. I won't attempt to start a list here, since this won't lead anywhere. Just an example:
Consider the map $S^1 \to S^1$, $z \mapsto z^p$ which winds the circle $p$ times around itsself. On homology $H_1(-,K)$, this induces a $K$-linear map map $K \to K$ (after having fixed the canonical generators of $H_1(S^1,K)$). But which one? It turns out that this is $x \mapsto px$. Actually this is an easy example, since one can verify this guess immediately from the definitions. But for more complicated spaces and maps, this kind of writing down a linear map on homology, cohomology or homotopy groups becomes terribly difficult. Often it isn't enough to track down the definitions, but rather one has to use sophisticated tricks. And this becomes even more advanced when all these vector spaces are pieces of a spectral sequence, whose differentials you would like to understand.
Another, more elementary example: It is known that $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C} \times \mathbb{C}$. Now consider the isomorphism $\mathbb{C} \times \mathbb{C} \to \mathbb{C} \times \mathbb{C}$ which swaps the entries, $(a,b) \mapsto (b,a)$. What do you think this map corresponds to on the tensor product $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$? Is it really the one which swaps the tensor factors? It turns out that it corresponds to $\mathrm{id} \otimes c$ (or $c \otimes \mathrm{id}$, depending on choices), where $c$ denotes the complex conjugation!
A: A finite-dimensional vector space $V$ is isomorphic to its dual $V^*$, but the isomorphism depends on the choice of a basis in $V$.
However, there is a canonical isomorphism between $V$ and its double dual $V^{**}=(V^*)^*$ given by $v \mapsto (f \mapsto f(v))$, which does not depend on a basis.
