If a vector space doesn't have an inner-product defined on it, how do the components of a vector change under the change of basis?
If $V$ is a vector space which doesn't have an inner-product defined on it. $\{e_{i = 1,2, \ldots}\} \ \text{s.t } e_i \in V$ form a basis $B_1$ of $V$ and $\{f_{j = 1,2, \ldots}\} \ \text{s.t } f_j \in V$ is another basis $B_2$ of V.
Any vector $v \in V$ can be expressed as $v = \sum_ic_ie_i = \sum_jd_jf_j $. In the absence of the innerproduct given $c_i \ , \ B_1, \ \text{and} \ B_2$ is it possible to find $d_j \ $ ?
The concepts of change of basis and inner product are independent of one another.
If a vector $v$ can be decomposed with respect to two bases as follows
$$v = \sum_ic_ie_i = \sum_jd_jf_j$$
then the way you relate the coefficients of those decompositions is by finding out how the elements of one basis can be expressed as a linear combination of elements of the other basis. Let's assume
$$e_i = \sum_j M_{ij}f_j$$
for some coefficients $M_{ij}$.
Then,
$$v = \sum_ic_ie_i = \sum_{i,j} c_i M_{ij} f_j$$
But since this is also equal to $\sum_j d_j f_j$, we can conclude that
$$\sum_j \left(\sum_i c_i M_{ij} - d_j\right) f_j = 0$$
Now, because the $f_j$ form a basis, the above equality is only true if all the coefficients are $0$ and thus
$$d_j = \sum_i c_i M_{ij}$$
which gives a relation between the coefficients of each decomposition of $v$.