Let a non trivial $\alpha\in \pi\_m(S^n)$ with a finite order $|\alpha|$. Write $G\_\alpha(S^m, S^n)$ for the path-component determined by $\alpha$ of the mapping space $G(S^m, S^n)$ of free maps. Next, take a natural number $k>1$ relatively prime to $|\alpha|$. Then, for a self-map $f\colon S^{n-1} \to S^{n-1}$ with degree $k$, consider the induced map $(\Sigma f)\_\* \colon G\_\alpha(S^m, S^n) \to G\_{k\alpha}(S^m, S^n)$.


Is it true that $(\Sigma f)\_\*$ is a homotopy equivalence?



Path component of the mapping spaces $G(S^m, S^n)$
This is not true. The short version is that even though $\alpha$ has order relatively prime to $k$, it's still a whole path component of $G(S^m,S^n)$. The homotopy groups of $G\_\alpha(S^m,S^n)$ then still contain a lot of $k$-torsion that tends to be killed by this degree-$k$ map.
$\require{AMScd}$




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Pick basepoints $p \in S^m$ and $q \in S^n$ (fixed by $\Sigma f$). Evaluation at $p$ determines a map of fibration sequences
$$
\begin{CD}
(\Omega^m S^n)\_\alpha @>>> G\_\alpha(S^m,S^n) @>>> S^n\\
@V\Omega^m \Sigma f VV @V (\Sigma f)\_\* VV @VV\Sigma f V\\
(\Omega^m S^n)\_{k \alpha} @>>> G\_{k\alpha}(S^m,S^n) @>>> S^n\\
\end{CD}
$$
where $(\Omega^m S^n)\_\alpha$ is the path component of the chosen element $\alpha$. In particular, the long exact sequence on homotopy groups implies that $(\Omega^m S^n)\_\alpha \to G\_\alpha(S^m,S^n)$ is an isomorphism on homotopy groups in degrees less than $(n-1)$. Therefore, it will suffice to find an example where this map on homotopy groups is not an isomorphism in this range.


For any basepoint-preserving map $g: S^n \to S^n$, we also have a commutative diagram
$$
\begin{CD}
(\Omega^m S^n)\_0 @>{\Omega^m g}>> (\Omega^m S^n)\_0\\
@V {t\_\alpha} V V @VV {t\_{g\_\* \alpha}} V\\
(\Omega^m S^n)\_\alpha @>>{\Omega^m g}> (\Omega^m S^n)\_{g\_\* \alpha}
\end{CD}
$$
where $t\_\beta: \Omega^m S^n \to \Omega^m S^n$ is the homotopy equivalence $x \mapsto \beta \ast x$ induced by loop multiplication. As a result, we can check if the map $\Omega^m (\Sigma f)$ determines an isomorphism $\pi\_\*(\Omega^m S^n)\_\alpha \to \pi\_\* (\Omega^m S^n)\_{k\alpha}$ by checking if it induces an isomorphism at the zero component. But at the zero component, we have a natural isomorphism
$$
\pi\_d (\Omega^m S^n)\_0 \cong \pi\_{d+m} S^n
$$
for any $d > 0$.


Therefore, we need to find:


* a nontrivial element $\alpha \in \pi\_m S^n$ of finite order,
* a $k > 1$ relatively prime to the order of $\alpha$, and
* a $d$ with $0 < d < n-1$ such that a degree $k$ map $S^n \to S^n$ does not induce an isomorphism on $\pi\_{d+m} S^n$.


As one way to make this easier on ourselves, if $d+m \leq 2n-2$, then $\pi\_{d+m} S^n$ is in the stable range (up to $\pi\_{2n-2}$), and a degree $k$ map $S^n \to S^n$ induces multiplication by $k$ on $\pi\_{d+m}$.




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Let's take $n \geq 5$ and $m = n+1$. Choose $\alpha$ be the Hopf element $\eta \in \pi\_{n+1} S^n$. Since $n > 2$, this element has order $2$, so we can pick $k=3$ (hence $\eta = 3\eta$). The above isomorphisms say
$$
\pi\_2 G\_\eta(S^{n+1},S^n) \cong \pi\_2 (\Omega^{n+1} S^n)\_\eta \cong \pi\_{n+3} S^n.
$$
This is in the stable range because $n \geq 5$, and so a degree $3$ map on $S^n$ induces multiplication-by-$3$ on $\pi\_{n+3} S^n$. Finally, this group is isomorphic to $\mathbb{Z}/24$ by classical calculations of stable homotopy groups of spheres, so multiplication-by-3 is not an isomorphism on it.


