I need to determinate if $\langle r\rangle$ is characteristic in $D_n = \langle r \rangle_n \rtimes \langle s \rangle_2$. This is trivial if I use the result that every cyclic group is characteristic (ALERT:This is wrong, see the comments to see why), but I found an automorphism $\alpha$ in $D_n$ where $\alpha(\langle r\rangle)\neq \langle r\rangle $.
In the special case $D_3$ I defined $\alpha$ to satisfy $\alpha(r)=s$ and $\alpha(s)=s$. I derived the value of $\alpha$ of the rest of elements from this two, and it looks like a valid automorphism. And of course the problem is that $\alpha(\langle r\rangle)= \{ 1,s \} \neq \langle r\rangle $
On the other hand I know that every morphism must satisfy that $|\alpha(r)|$ divides $|s|$, but this is not the case.
So, what is failing in $\alpha$?
I assume that by $D_n$ you mean the dihedral group of order $2n$, that is, $$D_n = \Bigl\langle r,s\Bigm| r^n = s^2 = 1,\ sr=r^{-1}s\Bigr\rangle.$$
The statement you claim is false for $n=2$. If $n=1$, you have the cyclic group of order $2$; then $\langle r\rangle$ is the trivial subgroup, which is characteristic. However, when $n=2$, you get the Klein $4$-group, and none of its proper nontrivial subgroups are characteristic. The result does hold for $n\gt 2$.
Assume $n\geq 3$. Let $\alpha$ be an automorphismn of $D_n$; we want to show that $\alpha(\langle r\rangle) = \langle r\rangle$ It suffices to show that $\alpha(r)\in\langle r\rangle$, since $r$ is a generator of the subgroup, so this implies $\alpha(\langle r\rangle)\subseteq\langle r\rangle$, and $\alpha(r)$ has order $|r|$, we get equality.
The elements of $D_n$ are: the elements of $\langle r\rangle$, which have order dividing $n$; and the elements of the form $r^is$, $0\leq i \lt n$, which have order $2$. The image of $r$ under $\alpha$ must have order $n\gt 2$, and hence must lie in $\langle r\rangle$, the only place where elements of order $n$ exist at all. Thus, $\alpha(r)\in\langle r\rangle$, showing that $\langle r\rangle$ is characteristic.
There are multiple errors in what you write. First, “characteristic” is not an intrinsic property (groups are not “characteristic”), it is an extrinsic, contextual property: subgroups are characteristic in other groups. So it makes no sense to say “cyclic groups are characteristics” (perhaps you are thinking that if $G$ is a cyclic group, then all of its subgroups are characteristic in $G$?). Second, it is false that cyclic subgroups are always characteristic, as the Klein $4$-group example shows. Thirdly, your $\alpha$ is not a group morphism; just because you write $\alpha(r)=s$ does not mean you have defined a group morphism. Recall that to define a morphism using the presentation, you must not only specify what the image of $r$ and $s$ are, you must also show that the images satisfy the defining relations. So to see whether $\alpha(r)=s$ and $\alpha(s)=s$ defines a group homomorphism from $D_n$ to itself, you would need to check that $\alpha(r)^n = \alpha(s)^2 = 1$, and $\alpha(s)\alpha(r) = \alpha(r)^{-1}\alpha(s)$. But the first one fails; so you have not defined a group homomorphism. Your $\alpha$ doesn’t work at all.