Question: My intuition says there must be difference because $\mathbb Z\subset\mathbb Q$ so there are elements in $\mathbb Q$ but not in $\mathbb Z$.

My question is about whether the following explanation true or not.

There is bijective mapping between $\mathbb Z$ and $\mathbb Q$ (because both are countable) so they are isomorph one another because of it their elements (elements of $\mathbb Z$ and of $\mathbb Q$ ) are just symbol so they are actually the same set with different symbol.

Answer: 
In the category of sets where the morphisms are arbitrary functions, isomorphisms are bijective functions. Since there exist bijections between the integers $\mathbb{Z}$ and $\mathbb{Q}$, they are isomorphic *in the category of sets*. Thus if the only structure you care about is the structure of a set, then you can treat $\mathbb{Z}$ and $\mathbb{Q}$ as the same object. In this setting, one can reasonably say that they are "equal".

However, we don't generally work in the category of sets. Typically, we want more structure than that. For example, in the *category of abelian groups*, the morphisms are group homomorphisms, and isomorphisms are group isomorphisms. A group isomorphism must be bijective, and preserve the group operation, i.e. it is a map $\varphi$ such that $\varphi(x+y) = \varphi(x) + \varphi(y)$ for all $x$ and $y$ in the domain.

In this setting, $\mathbb{Z}$ and $\mathbb{Q}$ are quite different, as there is no bijective map $\varphi : \mathbb{Z} \to \mathbb{Q}$ such that $\varphi(m + n) = \varphi(m) + \varphi(n)$ for all integers $m$ and $n$. To see this, suppose that $\varphi$ is a homomorphism and note that as soon as you know what $\varphi(1)$ is, you have completely determined what $\varphi$ does. If $\varphi(1) = r$, then $\varphi(\mathbb{Z})$ is the set of all integer multiples of $r$ in $\mathbb{Q}$. But $\frac{r}{2}$ is a rational number, and is not in the image of $\mathbb{Z}$, hence it is impossible for $\varphi$ to be surjective onto $\mathbb{Q}$.
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> **Theorem:** If $Y$ is a finite set, then any injection $X \to Y$ is an isomorphism (of sets) if and only if it is surjective.
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I posit that what your intuition is not about *cardinality*, but instead about is the notion:

* $f : \mathbb{Z} \to \mathbb{Q}$ is not surjective

However, most of your experience is probably about finite sets, and so your experience has mislead you to intuit "$X$ and $Y$ have the same cardinality" and "all injections $X \to Y$ are bijections" the same way.

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Another possibility of where your intuition has gone astray is

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> **Theorem:** If $Y$ is a finite set, then if any one morphism $X \to Y$ is an isomorphism, then *all* injections $X \to Y$ are isomorphisms.
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and your experience with finite sets has led you to intuit "same cardinality" with "*all* injections are bijections" the same.