Induced maps on homotopy groups of $SU(2) \rightarrow G$ We have $\pi_3(G) = \mathbb{Z}$ for all compact connected simple Lie groups, and we know that given a map $\phi: SU(2) \rightarrow G$, the induced map $\phi_{*} : \mathbb{Z} \rightarrow \mathbb{Z}$ on the $3^{rd}$ homotopy groups is given by the so-called Dynkin index. See here.
Let $\phi_{min}: SU(2) \rightarrow G$ denote the map with Dynkin index $1$.
(The labelling of the map with a subscript 'min' is because this map comes from the minimal nilpotent orbit of $\mathfrak{su}(2)$ in $\mathfrak{g}$, I am unsure if this is significant or not).
Anyway, my question is the following, can we say anything about the induced map $\phi_{{min *}}$ between homotopy groups of degrees $\geq 4$.
The specific case I am interested in is when $G = Sp(12)$ and calculating maps between $\pi_4$ and $\pi_5$. In this case we have:
$\pi_4(SU(2)) = \pi_5(SU(2)) = \mathbb{Z}_2$
$\pi_4(G) = \pi_5(G) = \mathbb{Z}_2$
Is it the case that the induced map $\phi_{min *} \equiv 1$ on both $\pi_4$ and $\pi_5$? Or is there more to be said?
 A: Before addressing your question, just a comment about the Dynkin index. While any map $\phi : SU(2) \to G$ induces a map $\phi_* : \pi_3(SU(2)) \to \pi_3(G)$, to obtain a map $\mathbb{Z} \to \mathbb{Z}$, you need to choose isomorphisms $\pi_3(SU(2)) \to \mathbb{Z}$ and $\pi_3(G) \to \mathbb{Z}$. The Dynkin index of $\phi$ will depend on the choice of these isomorphisms. However, since the only non-trivial isomorphism $\mathbb{Z} \to \mathbb{Z}$ is multiplication by $-1$, the Dynkin index of $\phi$ is well-defined up to sign. With this in mind, we see that $\phi_* : \pi_3(SU(2)) \to \pi_3(G)$ is an isomorphism if and only if there exist a choice of isomorphisms $\pi_3(SU(2)) \to \mathbb{Z}$ and $\pi_3(G) \to \mathbb{Z}$ such that the induced map $\mathbb{Z} \to \mathbb{Z}$ is multiplication by $1$.
Returning to the question at hand, first recall that $SU(2) \cong Sp(1)$. Second, note that $Sp(n)$ is the quaternionic unitary group, so it acts transitively on $S^{4n-1} \subset \mathbb{H}^n$ with stabiliser $Sp(n-1)$. Fixing $x = (0, \dots, 0, 1) \in \mathbb{H}^n$, we obtain a fiber bundle $Sp(n-1) \to Sp(n) \xrightarrow{p} S^{4n-1}$, where $p(A) = Ax$. Let $i_n : Sp(n-1) \to Sp(n)$ be the standard inclusion map $A \mapsto \begin{bmatrix} A & 0\\ 0 & 1\end{bmatrix}$; this is precisely the inclusion of $Sp(n-1)$ as $p^{-1}(x)$, the stabiliser of $(0, \dots, 0, 1) \in \mathbb{H}^n$. There is an associated long exact sequence in homotopy
$$\dots \to \pi_{k+1}(S^{4n-1}) \to \pi_k(Sp(n-1)) \xrightarrow{(i_n)_*} \pi_k(Sp(n)) \xrightarrow{p_*} \pi_k(S^{4n-1}) \to \pi_{k-1}(Sp(n-1)) \to \dots$$
It follows that $(i_n)_* : \pi_k(Sp(n-1)) \to \pi_k(Sp(n))$ is an isomorphism for $k \leq 4n - 3$.
Now let $i = i_n\circ\dots\circ i_2$, then $i : Sp(1) \to Sp(n)$ is the standard inclusion map $A \mapsto \begin{bmatrix} A & 0\\ 0 & I\end{bmatrix}$ and $i_* = (i_n)_*\circ\dots\circ(i_2)_*$.
For $m \geq 2$, we have $3 \leq 4m - 3$ so $(i_m)_* : \pi_3(Sp(m-1)) \to \pi_3(Sp(m))$ is an isomorphism, and hence so is $i_* : \pi_3(Sp(1)) \to \pi_3(Sp(n))$.
Likewise, for $m \geq 2$, we have $4, 5 \leq 4m - 3$ so $(i_m)_* : \pi_4(Sp(m-1)) \to \pi_4(Sp(m))$ and $(i_m)_* : \pi_5(Sp(m-1)) \to \pi_5(Sp(m))$ are isomorphisms, and hence so are $i_* : \pi_4(Sp(1)) \to \pi_4(Sp(n))$ and $i_* : \pi_5(Sp(1)) \to \pi_5(Sp(n))$.
