Let $SO(n)$ be the special orthogonal group. There is a self-diffeomorphism $\phi:SO(n)\to SO(n)$ taking an element $A$ to $A^{-1}$. I am interested in the induced automorphism $\phi_\ast:\pi_i(SO(n))\to \pi_i(SO(n))$. I am mostly interested in the values of $i$ in the stable range. In this range, Bott periodicity tells us that the homotopy groups are trivial, $\mathbb{Z}_2$, or $\mathbb{Z}$. The first two of these don't have any interesting automorphisms, so the question is really only interesting when the homotopy group is $\mathbb{Z}$, so the interesting version of my question is the following.
Let $i$ be in the the stable range for $SO(n)$ and suppose that $\pi_i(SO(n))\cong \mathbb{Z}$. Is $\phi_\ast(1)=1 $ or $-1$?
One computation I have done already is the case of $\pi_1(SO(2))$ where $\phi_\ast(1)=-1$. However, $i$ is not in the stable range, so this doesn't really count. Unfortunately, the first interesting version of the question involves a computation in $\pi_3(SO(5))$, and I don't really have any idea about how to think about this group.
I could imagine the answer depending on the value of $n$ and/or $i$. Sort of like how the antipodal map for spheres is homotopic to the identity in odd dimensions and not in homotopic to the identity in even dimensions.
Thinking through this a bit more, I think the following calculates the action on the inversion map on $\pi_3(\mathrm{SO}_5)$ (and indeed the method does not use anything about that group in particular, so the method should work for any compact Lie group $G$ for which $\pi_3(G) = \mathbb Z$).
For any group $G$, let $\iota_G\colon G \to G$ be the inversion map, $\iota_G(g)= g^{-1}$ for any $g \in G$. We wish to calculate the map $(\iota_{\mathrm{SO}_5})_*\colon \pi_3(\mathrm{SO}_5)\to \pi_3(\mathrm{SO}_5)$.
If $G=\mathrm{SO}_5$, then $\mathrm{Lie}(G) = \mathfrak g$ complexifies to the simple complex Lie algebra $\mathfrak{g}_{\mathbb C}$ with Weyl group of type $B_2=C_2$, i.e. the dihedral group $D_8$. You can recover $\mathfrak g$ from $\mathfrak g_{\mathbb C}$ the real Lie algebra as the real form on which the Killing form is negative definite.
The root system has 4 positive roots, and for each positive root, the root spaces $\mathfrak g_{\alpha}$ and $\mathfrak{g}_{-\alpha}$ generate a subalgebra $\mathfrak{sl}_{\alpha}$ which is isomorphic to $\mathfrak{sl}_2(\mathbb C)$. The intersection of $\mathfrak{sl}_{\alpha}$ with the compact real form of $\mathfrak{g}$ defines a Lie algebra homomorphism $i_\alpha\colon \mathfrak{su}_2 \to \mathfrak{so}_5$ with image $\mathfrak s_{\alpha}$. The homomorphism $i_\alpha$ then exponentiates to give a homomorphism of Lie groups $I_{\alpha} \colon \mathrm{SU}_2\to \mathrm{SO}_5$ which is a covering map onto its image, and hence the image $S_{\alpha}$ is isomorphic to $\mathrm{SU}_2$ or $\mathrm{SO}_3$.
Since $\mathrm{SU}_2$, as a topological space, is the $3$-sphere $S^3$, we obtain a map $(i_\alpha)_*\colon \pi_3(S^3) \to \pi_3(\mathrm{SO}_5)$. If this is nonzero $(i_{\alpha})_*$ then one can detect the action of the inversion map using it since for any group homomorphism $\phi\colon G\to H$ then $\phi\circ \iota_H = \iota_G \circ \phi$. It follows that $\phi_*\circ (\iota_{\text{SU}_2})_*= (\iota_{\mathrm{SO}_5})_*\circ\phi_*$
Now one can check that $(i_{\alpha})_*$ is nonzero by showing its image in $H_3(\mathrm{SO}_5)$ pairs nontrivially with an invariant 3-form. This 3-form on the group is is given by the 3-form $(X,Y,Z) \mapsto \langle X,[Y,Z]\rangle$ on $\mathfrak{so}_5$ where $[-.-]$ is the Lie bracket and $\langle -,-\rangle$ is the Killing form, an invariant symmetric bilinear form on $\mathrm{so}_5$.
Now if we identify $\mathrm{SU}_2$ with $\mathbb U$ the unit quaternions, $\mathbb U = \{h = a+bi+cj+dk\mid a^2+b^2+c^2+d^2=1\}$, then $h^{-1} = a-bi-cj-dk$, so that $\iota_{\mathbb U}\colon \mathbb U \to \mathbb U$ is the action of $D=\mathrm{diag}(1,-1,-1,-1)$ on $\mathbb R^4$ restricted to the $3$-sphere. Since $\det(D)=-1$, it is in the connected component containing the reflections, and hence its action on $H_3(S^3)$ is by $-1$, so that its action on $\pi_3(S^3)$ must therefore also be by $-1$.
Addendum
Since $S^1$ and $S^3$ are the only (connected) spheres which are Lie groups, the above strategy doesn't generalise in any obvious way to $\pi_k(\mathrm{SO}_n)$ with $k \neq 3$. That said, there is an obvious way to relate the homotopy groups of $\mathrm{SO}_n$ to the homotopy groups of spheres, because if $$ S^n = \{ \mathbf v \in \mathbb R^{n+1}: \|v\|=1\} $$ Then it is a homogeneous space for $\mathrm{SO}_{n+1}$ with the stabilizer of a vector $\mathbf v\in S^{n}$ being $\mathrm{SO}(H)$ where $H= (\mathbb R.\mathbf v)^{\perp}$. Since $\mathrm{SO}(H)\cong \mathrm{SO}_{n}$, the long exact sequence of the fibration
$$\require{AMScd} \begin{CD} \mathrm{SO}_n @>>> \mathrm{SO}_{n+1} \\ @. @VVV \\ @. S^n \end{CD}$$
allows one to study the homotopy groups of the special orthogonal groups inductively from those of the spheres (but obviously this is not trivial to do!)