Here's a simple example for $n=4$, namely a finite subgroup of $\mathrm{GL}_4(\mathbf{R})$ not contained in any finite subgroup of $\mathrm{GL}_4(\mathbf{R})$ generated by reflections.
Let $p$ be a prime $\ge 7$, and let $M$ be the $2\times 2$ matrix of order $p$ given by rotation by $2\pi/p$. Consider the group $G\subset\mathrm{GL}_4(\mathbf{R})$ generated by $M_1=\begin{pmatrix}M &0\\ 0 & I_2\end{pmatrix}$, $M_2=\begin{pmatrix}I_2 &0\\ 0 & M\end{pmatrix}$, $s=\begin{pmatrix}0 & I_2\\ I_2 & 0\end{pmatrix}$. (It has order $2p^2$, with an abelian subgroup of index 2 $H$ generated by $M_1,M_2$, and is isomorphic to the wreath product $C_p\wr C_2$ where $C_n$ is cyclic of order $n$.)
I claim that $G$ is not contained in a finite subgroup of $\mathrm{GL}_4(\mathbf{R})$ generated by reflections. There are two steps.
1) $G$ is not contained in a finite subgroup of $\mathrm{GL}_4(\mathbf{R})$ generated by reflexions and normalizing $H$. Indeed, $H\simeq C_p\times C_p$ fixes exactly 2 elements in the 2-Grassmanian of $\mathbf{R}^4$, namely the planes $x_3=x_4=0$ and $x_1=x_2=0$. Hence the normalizer $L$ of $H$ preserves this unordered pair of planes, and hence has a subgroup $L'$ of index (at most) 2 preserving these two subspaces; this index is actually 2 since $s$ belongs to this normalizer but exchanges these 2 subspaces. But $L\smallsetminus L'$ consists of elements of trace zero and hence contains no reflection. Therefore any subgroup of $L$ containing $G$ fails to be generated by reflections.
2) any finite subgroup of $\mathrm{GL}_4(\mathbf{R})$ containing $H$ normalizes $H$. This follows from the classification of subgroups of $\mathrm{SO}(4)$. Indeed, first observe that it follows from the classification of finite subgroups of $\mathrm{SO}(3)$ that in such a finite subgroup, any cyclic subgroup of order $p$ is normal (we use here that $p\neq 2,3,5$), since finite subgroups of $\mathrm{SO}(3)$ with an element of order $p$ are cyclic or dihedral. That cyclic subgroups of order $p$ are normal and $p$-Sylow in all finite subgroups containing them therefore remains true in the double cover $\mathrm{SU}(2)$. It follows that in any finite subgroup $F$ of $\mathrm{SU}(2)\times\mathrm{SU}(2)$, any subgroup of $F$ isomorphic to $C_p\times C_p$ is normal and $p$-Sylow in $F$; and in turn the same conclusion holds in $\mathrm{SO}(4)=(\mathrm{SU}(2)\times\mathrm{SU}(2))/C_2$, and in turn in $\mathrm{O(4)}$, and in turn in $\mathrm{GL}_4(\mathbf{R})$.
The conclusion follows from 1) and 2).
It is true that $a$ and $b$ do not commute, but that does not mean that the elements $ba,ba^2,ba^3,...ba^{n-1}$ are all distinct from the elements $ab,a^2b,a^3b,...a^{n-1}b$. Indeed, with the usual choice of generators for the dihedral group, we have $ba=a^{n-1}b$, and it follows that $ba^r=a^{n-r}b$ for all $r$ and so your last $n-1$ elements are all repeats of elements you already had.
Best Answer
A symmetry must map vertex 1 to some vertex $i$. Vertex 2 must be mapped to an ajacent vertex, $i+1$ or $i-1$. Once you know the images of 1 and 2, the other vertices are fixed. If 2 is mapped to $i+1$ you have the rotations. If 2 is mapped to $i-1$, you have the reflections (This might require some more arguments, at the elementary level at which this question is posed.)