$\DeclareMathOperator\Mod{Mod}\DeclareMathOperator\Homeo{Homeo}$Let $\Mod(M)=\pi_0(\Homeo(M))$ be the mapping class group of a manifold, possibly with boundary (I'm including the orientation reversing classes). Let $P$ be a $1$-sided, embedded $\mathbb{RP}^2$ in $\mathbb{RP}^3$. Its tubular neighborhood is denoted by $\nu(\mathbb{RP}^2)$ and it is a twisted $I-$bundle over $\mathbb{RP}^2$. The boundary $\partial \nu \mathbb{RP}^2$ is an embedded sphere, hence $\mathbb{RP}^3 = \nu(\mathbb{RP}^2) \cup_{S^2} \mathbb{B}^3$.

I know that $\nu(\mathbb{RP}^2) \subset \mathbb{RP}^3$ equals to the quotient $\left(S^2 \times I\middle)\right/_\sim$ with $(x,1)\sim(-x,1)$ thus $\nu(\mathbb{RP}^2) \setminus \mathbb{RP}^2 = S^2\times [0,1)$. I know that $\Mod(S^2)$= $\mathbb{Z}/2\mathbb{Z}= \Mod(\mathbb{B}^3)$ and $\Mod(\mathbb{RP}^2) = 0$ so I'm looking for a way to either extend the isotopy on $\mathbb{RP}^2$ to the whole $\nu(\mathbb{RP}^2)$, or extend the isotopy on the sphere $\partial \nu(\mathbb{RP}^2)$ into the interior. Is that possible?

Is there an easy, topological way to describe $\pi_0(\Homeo(\nu(\mathbb{RP}^2)))$? Alternatively, is there an easy, topological way to show that $\Mod(\mathbb{RP}^3)$ is the $2$-cyclic group?