Action of a degree $m$ map on $\pi_{n+1}S^n$ Let $\delta_m \colon S^n \to S^n$ be a map of degree $m$. Then $\delta_m$ induces a map $\pi_{n+1}S^n \to \pi_{n+1}S^n$ by $\delta_m[f]:=[\delta_m\circ f]$. We know that for $n>2$ the group $\pi_{n+1}S^n$ is isomorphic to $\mathbb Z_2$. 
The question is, what kind of map $\delta_m \colon \pi_{n+1}S^n \to \pi_{n+1}S^n$ do we get?
 A: This is to correct an error in Eric Wolfsey's post.
Actually, $\pi_3S^2\cong\mathbb{Z}$ is generated by the Hopf map $\eta=\eta_2$, and $[\iota_2,\iota_2]=2\cdot\eta$. You can see this since the Whitehead product $[\iota_2,\iota_2]$ generates the kernel of the suspension $\Sigma:\pi_3S^2\rightarrow\pi_4S^3\cong\mathbb{Z}_2\{\eta_3\}$, which is onto.
However, the quadratic nature of the Hopf map is well know, and indeed it does obey 
$$\delta^{(2)}_m\circ \eta\simeq m^2\cdot \eta_2.$$
That this is true is actually a slightly deeper result due to the fact that $\eta$ has Hopf invariant one.
Now write $\delta^{(n)}_m:S^n\rightarrow S^n$ for the degree $m$ map on $S^n$. For $n\geq 2$, this map is the $(n-1)$-fold suspension of the degree $m$ map on $S^{1}$, that is $\delta^{(n)}_m\simeq \Sigma^{n-1}\delta^{(1)}_m$, and for $n\geq 3$, the homotopy group $\pi_{n+1}S^n$ is generated by the suspension $\eta_n=\Sigma^{n-2}\eta$. 
Recall that the (reduced suspension) is the same as smashing with $S^1$, that is $\Sigma X\simeq S^1\wedge X$, for any based CW space $X$. Moreover, there are homeomorphisms $S^{n+m}\cong S^n\wedge S^m\cong S^m\wedge S^n$, and in particular $S^{n+1}\cong S^1\wedge S^n$. One thing to take into consideration, however, is the Milnor sign convention, which states that the homeomorphism $S^n\wedge S^m\cong S^m\wedge S^n$ has degree $(-1)^{nm}$.
We write $\iota_n=id_{S^n}:S^n\rightarrow S^n$ for the identity map. Then using the above facts we get a homotopy
$$\delta_m^{(n)}\circ\eta_n\simeq (\delta^{(1)}_m\wedge\iota_{n-1} )\circ(\iota_1\wedge \eta_{n-1})=(\iota_1\wedge \eta_{n-1})\circ(\delta^{(1)}_m\wedge\iota_{n})\simeq (-1)^{n}\cdot \eta_n\circ\delta^{(n)}_m.$$
However the right-hand side of this is just $=(-1)^nm\cdot \eta_n$. Since $\pi_{n+1}S^n\cong\mathbb{Z}_2$, the $(-1)^n$ factor is inconsequential, and we have
$$\delta_m^{(n)}\circ \eta_n=m\cdot \eta_n=\begin{cases}\eta_n&\text{$m$ odd}\\0&\text{$m$ even}.\end{cases}$$
Hence for $n\geq 3$, we have $\delta^{(n)}_*=\times m:\pi_{n+1}S^n\rightarrow\pi_{n+1}S^n$.
This actually agrees with Eric's previous answer, since $m^2\equiv m\mod 2$, but corrects his slightly faulty reasoning for the $n=2$ case.
A: For $n=2$, a nonzero element of $\pi_3(S^2)\cong\mathbb{Z}$ is  the Whitehead square $[\iota,\iota]$ where $\iota\in\pi_2(S^2)$ is the canonical generator (in fact, $[\iota,\iota]$ is twice a generator of $\pi_3(S^2)$).  Since the Whitehead product is bilinear, a degree $m$ map on $S^2$ maps $[\iota,\iota]$ to $[m\iota,m\iota]=m^2[\iota,\iota]$.  It follows that a degree $m$ map acts on $\pi_3(S^2)$ by multiplication by $m^2$.
For $n>2$, the $(n-2)$-fold suspension $\pi_3(S^2)\to \pi_{n+1}(S^n)$ is surjective.  Since a degree $m$ map on $S^n$ is just an $(n-2)$-fold suspension of a degree $m$ map on $S^2$, it acts on $\pi_{n+1}(S^n)$ by multiplication by $m^2$.  In other words, it acts by the identity if $m$ is odd and by $0$ if $m$ is even.
More generally, in the stable range of homotopy groups of spheres, a degree $m$ map acts by multiplication by $m$.  Indeed, if $\pi_k(S^n)$ is in the stable range, then $S^n\to \Omega S^{n+1}$ induces an isomorphism on $\pi_k$.  Now observe that by Eckmann-Hilton, the group operation on $\pi_k(\Omega S^{n+1})$ coincides with the group operation induced by concatenation of loops in $\Omega S^{n+1}$.  But now observe that the composition $S^n\stackrel{\delta_m}\to S^n\to \Omega S^{n+1}$ is homotopic to the composition $S^n\to \Omega S^{n+1}\stackrel{\mu_m}\to\Omega S^{n+1}$ where $\mu_m$ takes a loop to its $m$-fold concatenation with itself (this is again by Eckmann-Hilton: there are two operations on $[S^n,\Omega S^{n+1}]$, one from the cogroup structure on $S^n$ and one from the group structure on $\Omega S^{n+1}$, and they must be equal).  This means that the map induced by $\delta_m$ on $\pi_k(S^n)\cong \pi_k(\Omega S^{k+1})$ is multiplication by $m$.
(Even more generally, composition is bilinear in the stable homotopy category.  Here we are using that the composition of a map $S^k\to S^n$ with the sum of $m$ copies of the identity map $S^n\to S^n$ is homotopic to the sum of $m$ copies of our map $S^k\to S^n$, which is a special case of that bilinearity.)
