**Added 9/7/16:**
I just got access to the paper:

James, I. M. On the iterated suspension. Quart. J. Math., Oxford Ser. (2) 5, (1954). 1–10

which is an explicit reference to Greg's questions on the level of homotopy groups. See section 2.

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${}^\dagger$**Added 9/2/16:** I realize now that what I wrote below doesn't answer Question (1). It supplies an argument for Question (2).

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The answer is yes.$^\dagger$

Let $G_n$ be the topological monoid of self equivalences of $S^{n-1}$,
not necessarily preserving the basepoint.

Let $F_n$ be the topological monoid of self equivalences of $S^n$,
preserving the basepoint.

Unreduced suspension defines a homomorphism $G_n \to F_n$.

Forgetting the basepoint defines a homomorphism $F_n \to G_{n+1}$.

In effect, you are asking whether the map
$$
O(n+1)/O(n) = S^n = G_{n+1}/F_n \to F_{n+1}/F_n
$$
is about
$(2n)$-connected.

Since the diagram
$$
\require{AMScd}
\begin{CD}
G_{n+1} @>>> G_{n+1}/F_n \\
@VVV @VVV \\
F_{n+1} @>>> F_{n+1}/F_n
\end{CD}
$$
is $\infty$-cartesian it suffices to show that the map $G_{n+1} \to F_{n+1}$ is about
$(2n)$-connected.

Think of $G_{n+1}$ as a set of components of the based mapping space
$F_\ast(S^{n}_+,S^{n})$ and think of $F_{n+1}$ as given similarly
by components of $\Omega^{n+1}S^{n+1} = F_\ast(S^{n+1},S^{n+1}) \cong
F_\ast(S^{n},\Omega\Sigma S^{n+1})$. Then we have an $\infty$-cartesian square
$$
\begin{CD}
G_{n+1} @>>> F_{n+1} \\
@V\cap VV @VV\cap V\\
F_\ast(S^{n}_+,S^{n}) @>>E^{\sharp}> \Omega^{n+1}S^{n+1}\, .
\end{CD}
$$
Consequently, we need to understand the homotopy fibers of the lower horizontal map, labeled $E^\sharp$, taken at the identity $1$.

Consider the commutative diagram
$$
\begin{CD}
\text{"?"} @>>> F_\ast(S^{n}_+,S^{n}) @>E^\sharp>> \Omega^{n+1}S^{n+1} \\
@VVV @VV E V @| \\
\Omega \Sigma S^{n} @> i>> F_\ast(S^{n}_+,\Omega \Sigma S^{n}) @>R>> \Omega^{n+1}S^{n+1} \\
@V (a) VV @VV HV @VVV \\
F_\ast(S^{n}_+,\Omega \Sigma S^{2n}) @= F_\ast(S^{n}_+,\Omega \Sigma S^{2n}) @>>> \ast
\end{CD}
$$
where the horizontal maps are fiber sequences, with the homotopy fibers
of the first two rows taken at $1$. We need to identify "?" in the $2n$ range (more precisely, we need to show that it is about $(2n)$-connected.

The middle column is a fiber sequence
in degrees up to approximately $2n$ (to get an actual fiber sequence, we only need to replace the first term by $F_\ast(S^n_+,\text{fib}(H))$; this works since the map
$S^n \to \text{fib}(H)$ is about $(3n)$-connected).
The map $E^\sharp $ is given by reduced suspension $E$
restricted to $S^{n+1} \subset \Sigma(S^n_+)$ (the map $R$ is defined by restriction).

The fiber of $R$ is given by
$\Omega \Sigma S^n$ by use of the cofiber sequence of based spaces
$S^{n+1}\to \Sigma (S^n_+) \to S^1$. However, we have to remember that the inclusion of the fiber
picks up the identity map of $S^{n+1}$ by means of the adjunction.
The map $H$ is the James-Hopf invariant.

The map $(a)$
is defined by the composition.

**Discussion.** Note that both the source and target of $(a)$ have the homotopy type of $S^n$ in degrees $< 2n$ approximately. At first glance one might think this map is trivial, as it would be
if we had taken fibers at the trivial map rather than at 1. But
since to define it we've implicitly used an equivalence $\Sigma (S^n_+) \simeq S^{n+1} \vee S^1$,
which doesn't desuspend, this will make the map (a) nontrivial.

The desired connectivity statement about "?" is a direct consequence of the following claim.

**Claim:** The map (a) has degree one on $\pi_n$ (which is the same as its degree on $H_n$).

Here's a sketch of the argument proving the claim.

Consider the composition
$$
\begin{CD}
S^n @>j>> \Omega\Sigma S^n @>i >> F_\ast(S^n_+,\Omega\Sigma S^n) = F(S^n,\Omega\Sigma S^n)
\end{CD}
$$
where $j$ is adjoint to the identity map. The map $j$ represents a generator of $\pi_n$
of $\Omega\Sigma S^n$.

The adjoint of the composite $i\circ j$ is a map
$$
g: S^n \times S^n\to \Omega \Sigma S^n\, .
$$
By the definition of $i$, the map $g$ is given by loop multiplication of the
following two maps: $(x,y) \mapsto (t\mapsto x \wedge t)$ and $(x,y) \mapsto (t\mapsto y \wedge t)$.

Let $J(X)$ be the James construction on a (nice) based space.
Then $J(X) \to \Omega\Sigma X$
is a weak equivalence, where the map is induced from the inclusion $X\to \Omega \Sigma X$
using the multiplicative structure on $\Omega \Sigma X$ (strictly speaking, we should replace loops by Moore loops to get the map). Explicitly, the map sends a word
$(x_1,\dots, x_j)$ to the product $\prod_i \gamma_i$
where $\gamma_i(t) := t\wedge x_i$. Let $J_2(X) = X \cup (X \times X)$ be filtration two of the James construction.

Then the map $g$ factors up to homotopy through the map
$$
h: S^n \times S^n\to J_2(S^n)
$$
given by
$h(x,y) = (x,y)$.

The James-Hopf invariant $H: J(X) \to J(X\wedge X)$
sits in a homotopy commuting diagram
$$
\begin{CD}
J_2(X) @> q >> J_2(X)/X = X\wedge X \\
@V\cap VV @VV\cap V \\
J(X) @>>H > J(X\wedge X)
\end{CD}
$$
where $q$ is the quotient map.
If $X$ is $r$-connected then the vertical maps are at least $(3r)$-connected.

It follows that the degree of the map $(a)$ on $\pi_n$
is identified with the degree of the composition
$$
\begin{CD}
S^n \times S^n @>g >> J_2(S^n) @>q>> J_2(S^n)/S^n = S^n \wedge S^n
\end{CD}
$$
in homology in dimension $2n$. But this is just the evident quotient map
which has degree $+1$.