Is $ (\mathbb N^{<\mathbb{N}}, \leq_{pref})$ isomorphic with $ (\mathbb Q^{<\mathbb{N}}, \leq_{pref})$? Let $\mathbb N^{<\mathbb{N}} $ be a set of finite sequences of natural numbers, and $\mathbb Q^{<\mathbb{N}}$ - a set of finite sequences of rational numbers. 
For a finite sequence $(a_n)$, $|(a_n)|$ stands for its length (a number of elements). We define a prefix order on both sets:
$(a_n) \leq_{pref} (b_n)$ if and only if for any $n \leq |(a_n)|, a_n=b_n$.
Check if $ (\mathbb N^{<\mathbb{N}}, \leq_{pref})$ is isomorphic with  $ (\mathbb Q^{<\mathbb{N}}, \leq_{pref})$.
I can see that those two sequences are isomorphic, but I have no idea how to prove it. I'm afraid that stating that $\mathbb{N}$ is equinumerous to $\mathbb Q$ is not enough. Can you give me some hints how to solve it? 
 A: There are a few typos in your question. I suspect you know this, but just to clarify: $\mathbb{N}^{<\mathbb{N}}$ is the set of all finite sequences of naturals, $\mathbb{Q}^{<\mathbb{N}}$ is the set of all finite sequences of rationals, and neither is a sequence (rather, each is a set of sequences).
It turns out that the countability of $\mathbb{Q}$ is all that is needed. Specifically, suppose $X$ and $Y$ are two sets with the same cardinality; then it turns out that $(X^{<\mathbb{N}}, \le_{pref})$ and $(Y^{<\mathbb{N}}, \le_{pref})$ are isomorphic.
The proof of this more general fact is no harder than the proof of the particular instance you're interested in, so let's work in this more general setting (I think it will make things easier to think about, actually, since you won't be tripped up by the question of exactly what sorts of symbols we're using in our sequences).
Saying "$X$ and $Y$ have the same cardinality" is the same as saying "there is a bijection $f: X\rightarrow Y$." Now:


*

*Do you see a way to use $f$ to turn a sequence of elements of $X$ into a sequence of elements of $Y$? HINT: if the first term of the $X$-sequence is $x$, what should the first term of the corresponding $Y$-sequence be?

*Can you show that this in fact gives an isomorphism $i_f$ between $(X^{<\mathbb{N}}, \le_{pref})$ and $(Y^{<\mathbb{N}}, \le_{pref})$?

Note in general that there won't be a unique bijection $f$ between $X$ and $Y$ (indeed, the only time we'll have a unique bijection is when $X$ and $Y$ have at most one element!); so there won't be a unique isomorphism from $(X^{<\mathbb{N}}, \le_{pref})$ to $(Y^{<\mathbb{N}}, \le_{pref})$. Similarly, since (as long as $X$ has more than one element) there are many (= more than one :P) different self-bijections of $X$, $(X^{<\mathbb{N}}, \le_{pref})$ will have lots of automorphisms. 
Finally, as a more advanced exercise: can you show that (as long as $X$ and $Y$ have more than one element) if $X$ and $Y$ have the same cardinality, then there is an isomorphism from $(X^{<\mathbb{N}}, \le_{pref})$ to $(Y^{<\mathbb{N}}, \le_{pref})$ which is not of the form $i_f$ for any $f: X\rightarrow Y$ (that is, which isn't induced by a bijection from $X$ to $Y$)? 
