I'm considering a more generic problem, allowing points to lie on lines containing edges. Let use the $\mathbf p_i$ for the pentagon vertices and $\mathbf q_i$ for the rational points. The pentagon is fixed by its center $\mathbf c$ and $\mathbf r = \mathbf p_1 - \mathbf c$. Let $R$ be the rotation matrix by $\frac{2\pi}{5}$. Also let $T = R^{-1} = R^\top$.

The vertices of the pentagon are given by $$ \mathbf p_i = \mathbf c + R^i \mathbf r, \quad i = 0,\dots,4. $$ Index $i$ in $\mathbf p_i$ is considered modulo 5.

The points $\mathbf q_i$ can be represented as $$ \mathbf q_i = (1 - \lambda_i)\mathbf p_i + \lambda_i \mathbf p_{i+1} = \mathbf c + R^i((1 - \lambda_i) I + \lambda_i R) \mathbf r, \quad \lambda_i \in \mathbb R. $$ We do not restrict $\lambda_i \in [0, 1]$, so $\mathbf q_i$ can be anywhere on the line passing through $\mathbf p_i$ and $\mathbf p_{i+1}$.

Let's consider auxiliary points $\mathbf a_i$ defined as $$ \mathbf a_i = T^i (\mathbf q_i - \mathbf c). $$ This points satisfy $\mathbf a_i = (1 - \lambda_i) \mathbf r + \lambda_i R \mathbf r$ so they all lie in some line $\ell = \operatorname{aff}(\mathbf r, R \mathbf r)$.

Let's construct such linear combinations of $\mathbf a_i$ so they are free of $\mathbf c$. This basically reduces to finding trivial combinations $\sum_{k=0}^4 \alpha_k T^k = O$. It's easy to see that columns of the linear combination vanish at the same time, so the equality reduces to $$ \sum_{k=0}^4 \alpha_k \cos \frac{2\pi k}{5} = 0\\ \sum_{k=0}^4 \alpha_k \sin \frac{2\pi k}{5} = 0 $$ There are three linearly independent solutions to the system, forming the following null space matrix $$ \begin{pmatrix} \phi \\ -1 & \phi\\ \phi & -1 & \phi\\ & \phi & -1\\ && \phi \end{pmatrix}. $$ Here $\phi = \frac{1 + \sqrt{5}}{2}$, the golden ratio. It is easy to verify that $$\phi T^{i+1} - T^i + \phi T^{i-1} = T^i (\phi (R + T) - I) = O.$$

The combinations that are free from $\mathbf c$ are: $$ \mathbf b_i = \phi \mathbf a_{i-1} - \mathbf a_i + \phi \mathbf a_{i+1} = \phi T^{i-1}\mathbf q_{i-1} - T^i\mathbf q_i + \phi T^{i+1}\mathbf q_{i+1}, \quad i = 1, 2, 3. $$ Points $\mathbf b_i$ also lie on a same line $\ell' = (2\phi - 1) \ell$. Therefore $$ (\mathbf b_3 - \mathbf b_2) \vee (\mathbf b_2 - \mathbf b_1) = 0. $$ Here $\vee$ is the pseudoscalar (or skew) product.

By using the trigonometric form $T^k = \cos \frac{2\pi k}{5} I + \sin \frac{2\pi k}{5} J$ and performing some lengthy computation, one can obtain $$ \frac {4}{\phi^2} (\mathbf b_3 - \mathbf b_2) \vee (\mathbf b_2 - \mathbf b_1) = \\ = \left(\sum_{i=0}^4 \mathbf q_i \vee \mathbf q_{i-1}\right) (\sqrt{5} - 1) + \left(\sum_{i=0}^4 \mathbf q_i \vee \mathbf q_{i-2}\right) (\sqrt{5} + 3) + \\ + \left(\sum_{i=0}^4 \mathbf q_i \cdot (\mathbf q_{i-1}-\mathbf q_{i-2})\right) \sqrt{10 + 2\sqrt{5}}. $$ The indices are considered modulo 5.

Now let $$ A = \sum_{i=0}^4 \mathbf q_i \vee \mathbf q_{i-1},\\ B = \sum_{i=0}^4 \mathbf q_i \vee \mathbf q_{i-2},\\ C = \sum_{i=0}^4 \mathbf q_i \cdot (\mathbf q_{i-1}-\mathbf q_{i-2}),\\ D = A + B,\\ E = 3B - A. $$ Note that $A, B, C, D, E \in \mathbb Q$.

We finally have a necessary condition $$ D \sqrt{5} + E + C \sqrt{10 + 2\sqrt{5}} = 0. $$ It is easy to show that there is no nontrivial solution: $$ 5D^2 + E^2 + 2\sqrt{5}DE = C^2 (10 + 2\sqrt{5})\\ 5D^2 + E^2 - 10C^2 = 2(C^2 - DE)\sqrt{5}\\ C^2 - DE = 0\\ 5D^2 + E^2 = 10C^2 = 10DE\\ (E - 5D)^2 = 20D^2 \\ E = (5 \pm 2\sqrt{5})D\\ C = D = E = 0. $$ So the only possibility is $A = B = 0$. But $A$ is twice the oriented area of the polygon formed by $\mathbf q_{0},\dots,\mathbf q_{4}$. Similarly, $B$ is twice the area of the star-shaped polygon with permuted vertices $\mathbf q_0, \mathbf q_2, \mathbf q_4, \mathbf q_1, \mathbf q_3$.

This surely can't happen when $\mathbf q_{i}$ lie on the sides, since then polygon $\mathbf q_0, \mathbf q_1, \mathbf q_2, \mathbf q_3, \mathbf q_4$ is convex and nonempty.

I'm considering a more generic problem, allowing points to lie on lines containing edges. Let use the $\mathbf p_i$ for the pentagon vertices and $\mathbf q_i$ for the rational points. The pentagon is fixed by its center $\mathbf c$ and $\mathbf r = \mathbf p_1 - \mathbf c$. Let $R$ be the rotation matrix by $\frac{2\pi}{5}$. Also let $T = R^{-1} = R^\top$.

The vertices of the pentagon are given by $$ \mathbf p_i = \mathbf c + R^i \mathbf r, \quad i = 0,\dots,4. $$ Index $i$ in $\mathbf p_i$ is considered modulo 5.

The points $\mathbf q_i$ can be represented as $$ \mathbf q_i = (1 - \lambda_i)\mathbf p_i + \lambda_i \mathbf p_{i+1} = \mathbf c + R^i((1 - \lambda_i) I + \lambda_i R) \mathbf r, \quad \lambda_i \in \mathbb R. $$ We do not restrict $\lambda_i \in [0, 1]$, so $\mathbf q_i$ can be anywhere on the line passing through $\mathbf p_i$ and $\mathbf p_{i+1}$.

Let's consider auxiliary points $\mathbf a_i$ defined as $$ \mathbf a_i = T^i (\mathbf q_i - \mathbf c). $$ This points satisfy $\mathbf a_i = (1 - \lambda_i) \mathbf r + \lambda_i R \mathbf r$ so they all lie in some line $\ell = \operatorname{aff}(\mathbf r, R \mathbf r)$.

Let's construct such linear combinations of $\mathbf a_i$ so they are free of $\mathbf c$. This basically reduces to finding trivial combinations $\sum_{k=0}^4 \alpha_k T^k = O$. It's easy to see that columns of the linear combination vanish at the same time, so the equality reduces to $$ \sum_{k=0}^4 \alpha_k \cos \frac{2\pi k}{5} = 0\\ \sum_{k=0}^4 \alpha_k \sin \frac{2\pi k}{5} = 0 $$ There are three linearly independent solutions to the system, forming the following null space matrix $$ \begin{pmatrix} \phi \\ -1 & \phi\\ \phi & -1 & \phi\\ & \phi & -1\\ && \phi \end{pmatrix}. $$ Here $\phi = \frac{1 + \sqrt{5}}{2}$, the golden ratio. It is easy to verify that $$\phi T^{i+1} - T^i + \phi T^{i-1} = T^i (\phi (R + T) - I) = O.$$

The combinations that are free from $\mathbf c$ are: $$ \mathbf b_i = \phi \mathbf a_{i-1} - \mathbf a_i + \phi \mathbf a_{i+1} = \phi T^{i-1}\mathbf q_{i-1} - T^i\mathbf q_i + \phi T^{i+1}\mathbf q_{i+1}, \quad i = 1, 2, 3. $$ Points $\mathbf b_i$ also lie on a same line $\ell' = (2\phi - 1) \ell$. Therefore $$ (\mathbf b_3 - \mathbf b_2) \vee (\mathbf b_2 - \mathbf b_1) = 0. $$ Here $\vee$ is the pseudoscalar (or skew) product.

By using the trigonometric form $T^k = \cos \frac{2\pi k}{5} I + \sin \frac{2\pi k}{5} J$ and performing some lengthy computation, one can obtain $$ \frac {4}{\phi^2} (\mathbf b_3 - \mathbf b_2) \vee (\mathbf b_2 - \mathbf b_1) = \\ = \left(\sum_{i=0}^4 \mathbf q_i \vee \mathbf q_{i-1}\right) (\sqrt{5} - 1) + \left(\sum_{i=0}^4 \mathbf q_i \vee \mathbf q_{i-2}\right) (\sqrt{5} + 3) + \\ + \left(\sum_{i=0}^4 \mathbf q_i \cdot (\mathbf q_{i-1}-\mathbf q_{i-2})\right) \sqrt{10 + 2\sqrt{5}}. $$ The indices are considered modulo 5.

Now let $$ A = \sum_{i=0}^4 \mathbf q_i \vee \mathbf q_{i-1},\\ B = \sum_{i=0}^4 \mathbf q_i \vee \mathbf q_{i-2},\\ C = \sum_{i=0}^4 \mathbf q_i \cdot (\mathbf q_{i-1}-\mathbf q_{i-2}),\\ D = A + B,\\ E = 3B - A. $$ Note that $A, B, C, D, E \in \mathbb Q$.

We finally have a necessary condition $$ D \sqrt{5} + E + C \sqrt{10 + 2\sqrt{5}} = 0. $$ It is easy to show that there is no nontrivial solution: $$ 5D^2 + E^2 + 2\sqrt{5}DE = C^2 (10 + 2\sqrt{5})\\ 5D^2 + E^2 - 10C^2 = 2(C^2 - DE)\sqrt{5}\\ C^2 - DE = 0\\ 5D^2 + E^2 = 10C^2 = 10DE\\ (E - 5D)^2 = 20D^2 \\ E = (5 \pm 2\sqrt{5})D\\ C = D = E = 0. $$ So the only possibility is $A = B = 0$. But $A$ is twice the oriented area of the polygon formed by $\mathbf q_{0},\dots,\mathbf q_{4}$. Similarly, $B$ is twice the area of the star-shaped polygon with permuted vertices $\mathbf q_0, \mathbf q_2, \mathbf q_4, \mathbf q_1, \mathbf q_3$.

This surely can't happen when $\mathbf q_{i}$ lie on the sides, since then polygon $\mathbf q_0, \mathbf q_1, \mathbf q_2, \mathbf q_3, \mathbf q_4$ is convex and nonempty.

Update. However there are rational solutions to $A = B = C = 0$, for example $$ \mathbf q_0 = \left(\frac{1}{5},\frac{1}{2}\right), \mathbf q_1 = (1,2), \mathbf q_2 = \left(1,\frac{1}{2}\right), \mathbf q_3 = \left(2,\frac{1}{2}\right), \mathbf q_4 = \left(2,\frac{7}{6}\right). $$ This particular one was obtained by trial and error method by plugging 1 and 2 for unknowns and solving the remaining system.

Knowing $\mathbf q_i$ we can compute $\mathbf b_i$, obtaining the $\ell'$ line. Divinding it by $2\phi-1$ we get the $\ell$ line where $\mathbf a_i, \mathbf r$ and $R\mathbf r$ are located. The $\mathbf r$ can be found from the following linear system $$ ((2\phi-1) \mathbf r - \mathbf b_1) \vee (\mathbf b_2 - \mathbf b_1) = 0\\ ((2\phi-1) R\mathbf r - \mathbf b_1) \vee (\mathbf b_2 - \mathbf b_1) = 0 $$ Similarly $\mathbf c$ can be obtained as the solution to $$ ((2\phi-1) (\mathbf q_0 - \mathbf c) - \mathbf b_1) \vee (\mathbf b_2 - \mathbf b_1) = 0\\ ((2\phi-1) T (\mathbf q_0 - \mathbf c) - \mathbf b_1) \vee (\mathbf b_2 - \mathbf b_1) = 0 $$

The layout of the points and the pentagon is shown in the picture below: