The typical form of an isomorphism between a $K$-vector space and its coordinate space $K^n$ Let $V$ be an arbitrary vector space of finite dimension $n$ over
the field $K$. It is known that in that case 
$$
V\simeq K^{n}.
$$
The canonical isomorphism which achieves this is 
\begin{eqnarray*}
 & \phi:K^{n}\rightarrow V\\
 & \left(x_{1},\ldots,x_{n}\right)\mapsto x_{1}\vec{v}_{1}+\ldots+x_{n}\vec{v}_{n},
\end{eqnarray*}
where $\left(\vec{v}_{1},\ldots,\vec{v}_{n}\right)$ is a fixed basis
of $V$.
Of course this isn't the only isomorphism - different basis give different
isomorphisms. But may there also be others ? In other words: Given
an arbitrary isomorphism $\varphi:K^{n}\rightarrow V$ is there always
a basis $\left(\vec{u}_{1},\ldots,\vec{u}_{n}\right)$, such that
in this basis $\varphi$ looks like above, i.e. 
$$
\varphi\left(x_{1},\ldots,x_{n}\right)=x_{1}\vec{u}_{1}+\ldots+x_{n}\vec{u}_{n}\ ?
$$
If not, could you give an example ?

Yes, there is always such a basis - specifically, given an isomorphism $\phi:K^n\to V$, then letting
$$\vec{u}_i=\phi(0,\ldots,\underbrace{1}_{i\text{th place}},\ldots,0)$$
we have that 
$$\phi(x_1,\ldots,x_n)=x_1\phi(1,0,\ldots,0)+\cdots+x_n\phi(0,\ldots,0,1)=x_1\vec{u}_1+\cdots+x_n\vec{u}_n.$$
The set $\{\vec{u}_1,\ldots,\vec{u}_n\}$ is a basis because 
$$x_1\vec{u}_1+\cdots+x_n\vec{u}_n=\phi(x_1,\ldots,x_n)=0\in V$$
if and only if $(x_1,\ldots,x_n)=(0,\ldots,0)\in K^n$, because $\phi$ is an isomorphism.
