Yes!
The group $G$ should be $D_5\times \Bbb Z_2$, where $D_5$ is the dihedral group of order ten.
If $a$ is a rotation of $2\pi/5$ counterclockwise from a bird's-eye view, $b$ is a horizontal symmetry from that view, and $c$ is a flip about the centre, then we have
$$\begin{align}
bab&=a^{-1},\\
a^5&=e,\\
b^2&=e, \\
c^2&=e,\\
ac&=ca,\\
bc&=cb,
\end{align}$$
which give the presentation
$$\begin{align}
D_5\times \Bbb Z_2&\cong\langle a,b\mid a^5, b^2, bab=a^{-1}\rangle \times\langle c\mid c^2\rangle \\
&\cong\langle a,b, c\mid a^5, b^2, c^2, bab=a^{-1},ac=ca, bc=cb\rangle
\end{align}$$
by this standard result.
As we can see here (in particular, here, although it is very technical${}^\dagger$), we can use $\Bbb Z_2^2$ (i.e., $V_4$) and $\Bbb Z_5$, but we would have to use the semidirect product:
$$\boxed{G\cong \Bbb Z_5\rtimes\Bbb Z_2^2.}$$
Also, $G\cong D_{10}.$
$\dagger$: The notation $\Bbb Z_5\rtimes\Bbb Z_2^2$ has some ambiguity. Really, it should be
$$\color{red}{\boxed{G\cong\Bbb Z_5\rtimes_\varphi\Bbb Z_2^2,}}$$
where, if $\mathbb{Z}_5 = \langle x\mid x^5 \rangle$ and $\mathbb{Z}_2^2 = \langle y, z\mid y^2,z^2, yz=zy \rangle$, we have the homomorphism
$$
\varphi:\Bbb Z_2^2\to\Bbb Z_2\le {\rm Aut}(\Bbb Z_5)$$
with $\varphi(y)(x) = x^{-1}$ and $\varphi(z)(x) = x.
$ (This determines the whole map by the homomorphism property.)
