Top is the category of topological spaces and continuous maps simply by definition; topology typically deals with continuous maps, making this category the most important one, and thus by convention it's the one meant when saying "the category of topological spaces".
(aside: other conventions on what Top or "the category of topological spaces" stands for are far more likely to disagree on what the objects are, rather than the morphisms. e.g. to make the objects be merely the compactly generated Hausdorff spaces)
You can, of course, make all sorts of other categories. The category of topological spaces and open maps is a perfectly reasonable category to make; it's just less useful.
It takes a bit to get used to, but category theory rejects the mindset that mathematics is about objects, with the mappings between objects being a derived notion. Instead, you need to consider objects and mappings as equals -- or even to consider the objects superfluous.
On that last point, my favorite example of a category whose emphasis is on the morphisms is matrix algebra. The set of all matrices, with composition defined by multiplication, form a category. (with addition, you get an Abelian category) The objects of this category really play no role beyond bookkeeping to say which matrix products are defined.
(This category is, of course, equivalent to the category of fintie-dimensional vector spaces and linear maps)
The "structure" preserved is how the elements relate to each other under the operations of the algebraic structure. For example, for commutative rings, if $\ a = b^2 - c^2 = (b-c)(b+c)\ $ is a difference of squares then applying a ring hom $\,h\, :\, r\mapsto \bar r\,$ shows that it remains a difference of squares in the image ring, viz. $\ \bar a = \bar b^2-\bar c^2 = (\bar b - \bar c)(\bar b +\bar c),\,$ so this particular ring-theoretic structure is preserved. Similarly preserved are polynomial expressions, i.e. expressions composed of basic ring operations (addition, multiplication) and constants $\,0,1.\,$
For example, a root of a polynomial remains a root of the image of the polynomial. So we can prove that a polynomial has no roots by showing it has no roots in a simpler homomorphic image, e.g. a ring with size so small that we can easily test all of its elements to see if they are roots, e.g. common parity arguments: $\,f(x) = x(x\!+\!1)+2n\!+\!1\,$ has no integer roots since $\,x(x\!+\!1)\,$ is even, so adding $2n\!+\!1$ yields an odd hence $\ne 0;\,$ equivalently $\!\bmod 2\!:\ f(0)\equiv 1\equiv f(1),\,$ i.e. $\,f\,$ has no roots mod $\,2,\,$ hence it has no integer roots. This idea generalizes as follows
Parity Root Test $\ $ A polynomial $\,f(x)\,$ with integer coefficients
has no integer roots when its $\rm\,\color{#0a0}{constant\,\ coefficient}\,$ and $\,\rm\color{#c00}{coefficient\,\ sum}\,$ are both odd.
Proof $\ $ If so then $\ \color{#0a0}{f(0)} \equiv 1\equiv \color{#c00}{f(1)}\,\pmod 2,\ $ i.e. $\:f\:$ has no roots in $\,\Bbb Z/2 = $ integers mod $\,2,\,$ therefore $\,f\,$ has no integer roots. $\ $ QED
In the same way, we can often reduce problems to "smaller", simpler problems in modular images. Because homs preserve the ambient algebraic structure, as above, we can often deduce information about the original problems from information gleaned in the simpler modular images. Such problem solving by modular reduction is an algebraic way of "dividing and conquering".
Best Answer
They are exactly homomorphisms of vector spaces (preserving linear structures). $\mathbb{R}^n$ and $\mathbb{C}^n$ with inner products are not plain vector spaces, they are Euclidean/Hilbert spaces. So their homomorphism would need to preserve inner products also (hence orthogonal/unitary).
Regarding isomorphisms, there is a theorem that $f:V\to W$ is an isomorphism if and only if $\dim V=\dim W$. And in that view, we often study isomorphisms of the form $f:V\to V$. These are denoted by $\mathrm{Aut}(V)$, the automorphisms of $V$.