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By definition, two vector spaces being isomorphic means that if $h$ is their isomorphism, at least, the following is true

$$ah(x)=h(ax)$$

where $a$ is an element of the field and $x$ is a vector. This means that $a$ should be in both fields. But think about vector spaces whose fields are isomorphic, but not equal. Then the structure of the VS (vector space)is not changed and, hence, should be something like isomorphic. Then what property do these VS have?

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The map $h$ defines isomorphism of $2$ vector spaces $V, W$ (over fields $K$ and $L$ respectively) if it is:
(1) bijective as a map between sets.
(2) $K$-linear, meaning that $\forall a \in K\,\, \forall v \in V$ we have: $a\cdot h(v) = h(a\cdot v)$.

Second condition automatically requires $W$ to have a structure of a vector space over $K$, otherwise it is nonsense to talk about $a\cdot h(v)$.
This structure can be defined in an indirect way. For example, as you've said, we might know that $K$ and $L$ are isomorphic, choose isomorphism $\varphi$ and for $a\in K, w\in W$ define $a\cdot w:= \varphi(a) \cdot w$.
Another example is when $K \hookrightarrow L$, then we can look at $W$ as a vector space over $L$ and as a vector space over $K$.

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  • $\begingroup$ Is there a name for "isomorphism pair" $(\varphi, h)$ such that $\varphi(\alpha)\cdot h(v)=h(\alpha v)$? $\endgroup$ – Garmekain Nov 3 '18 at 17:59
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Given that two vector spaces are isomorphic iff there exists an isomorphism (i.e., a linear map with some additional requirements) between them, it does't make sense to ask yourself if two vector spaces on two different fields are isomorphic. Note that a linear map is, by definition, a function $ L $ from two $ \mathbb{K} $-vector spaces $ V $, $ W $ such that $ L(u+v)=Lu+Lv $ and $ Lcv=cLv $: for the second condition to be satisfied, it must be possible, inside $ W $, to multiply by a scalar of $ \mathbb{K} $; that's why we require the spaces to be over the same field. You can also say that two vector spaces are isomorphic iff they are isomorphic inside $ \mathbf{Vect}_\mathbb{K} $.

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  • $\begingroup$ There is no a predefined addition operator between two vectors of different vector spaces. $\endgroup$ – Garmekain Nov 3 '18 at 18:11
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    $\begingroup$ Of course, but where I said that? $\endgroup$ – user457568 Nov 3 '18 at 18:17

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