Dimension of $\mathbb{Q}$-vector space if a nonsingular linear transformation $T$ exists such that $T^{-1} = T^{2} + T$

We're given $V$ a finite dimensional vector space over $\mathbb{Q}$, $T$ a non-singular linear transformation of $V$ such that $T^{-1} = T^{2} + T$. The question has two parts. If I understand part (a), I should be able to get part (b), so right now I'm looking for a hint on part (a):

(a) Prove that the dimension of $V$ is divisible by 3. (b) Prove that if the dimension is exactly 3, then all such transformations $T$ are similar.

I'm trying to think of $T$ as a matrix. I can get $v = A^{3}v + A^{2}v$ for any $v$, but I don't know where to go from there.

(a) We can rewrite the given condition as $T^3+T^2-I=0$ and $x^3+x^2-1$ is irreducible over $\Bbb{Q}.$ Therefore this is the minimal polynomial $m_T(x)$ satisfies by $T.$

Since characteristic polynomial $C_T(x)$ and minimal polynomial has same zeros, here the only possibility is $C_T(x)=(m_T(x))^r.$ Also $\dim_{\Bbb{Q}}V=\deg(C_T(x))=3r.$

OR: we can note that $m_T(x)$ is the only invariant factor of $T.$ Hence $$V=\Bbb{Q(x)}/(m_T(x))\oplus\Bbb{Q(x)}/(m_T(x))\oplus\cdots \oplus\Bbb{Q(x)}/(m_T(x))$$ (finite direct sum) and further $\Bbb{Q(x)}/(m_T(x))=\langle 1, x, x^2\rangle_{\Bbb{Q}}$ implies $3|\dim_{\Bbb{Q}}V.$

(b) To answer this part, we can easily see that $C_T(x)=m_T(x).$ Therefore this define a unique Smith normal form $$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & x^3+x^2-1 \\ \end{pmatrix} $$ and unique Frobenius normal form $$ \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & -1 \\ \end{pmatrix} .$$ Hence the matrix representation of all such transformations has to be similar to the above second matrix.