Square Roots of Diffeomorhpisms of Manifold Conjugate?

Question

Is this known: given a smooth manifold $Q^n$, a diffeomorphism $f: Q \to Q$ that is isotopic to the identity, and two different "square roots of $f$", that is, $g_1: Q \to Q$ and $g_2: Q \to Q$ with $g_1 \ne g_2$, $g_1$ and $g_2$ both also diffeomorphisms and isotopic to the identity, and $g_1^2 = f = g_2^2$, is it necessarily the case that $g_1$ and $g_2$ are conjugate, that is, that there is a diffeomorphism $q: Q \to Q$ with $q \circ g_1 = g_2 \circ q$? (It may be the case that $Q$ is actually some kind of tangent bundle, $Q = TT\ldots TQ' = T^nQ'$, in which case we would want $q$ to be a bundle map, I think.)

(See also this post and this post.)

Answer 1

No, this is already quite false even if everything in sight is linear. Take $Q = \mathbb{R}^n$. Every linear map $T \in GL_n^{+}(\mathbb{R})$ of positive determinant gives a diffeomorphism of $Q$ isotopic to the identity. If $T$ is diagonalizable with positive real eigenvalues then generically $T$ has $2^n$ square roots which are generically not conjugate (conjugacy preserves the eigenvalues of a diffeomorphism at a fixed point), and the ones with positive determinant remain connected to the identity.

To be explicit, take $T = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$ and consider two of its square roots $\pm T$. In this case non-conjugacy is easy to see because the identity diffeomorphism is only conjugate to itself.