Continuous on the unit ball – odd on the unit sphere – does it have a fixed point? For $n\in\mathbb N$, let
\begin{align*}
B^n\equiv&\;\{\mathbf x\in\mathbb R^n\,|\,\lVert \mathbf x\rVert\leq 1\}\text{ and}\\
S^{n-1}\equiv&\;\{\mathbf x\in\mathbb R^n\,|\,\lVert \mathbf x\rVert= 1\}
\end{align*}
denote the unit ball and the unit sphere, respectively, where $\lVert\cdot\rVert$ is the standard Euclidean norm.
Suppose that $F:B^n\to\mathbb R^n$ is a continuous function satisfying $F(-\mathbf x)=-F(\mathbf x)$ for every $\mathbf x\in S^{n-1}$. That is, $F|_{S^{n-1}}$ is an odd function.

 Claim: The function $F$ has a fixed point, that is, some $\mathbf x^*\in B^n$ satisfying $F(\mathbf x^*)=\mathbf x^*$.

(The case $n=1$ is an immediate consequence of the intermediate-value theorem.)
(Disclosure: I spent the better half of this afternoon trying to construct a counterexample for $n=2$, in vain.)

Note: The function $F$ does not necessarily map into $B^n$, so Brouwer’s fixed-point theorem doesn’t apply directly. That said, upon defining the projection function $\mathsf p:\mathbb R^n\to B^n$ as
\begin{align*}
\mathsf p(\mathbf x)\equiv\frac{1}{\max\{\lVert\mathbf x\rVert,1\}}\mathbf x\quad\text{for each $\mathbf x\in\mathbb R^n$},
\end{align*}
the composite function $\mathsf p\circ F:B^n\to B^n$ is guaranteed to have a fixed point. Yet, it is unclear to me whether and how this observation helps prove that $F$ has a fixed point. (I would like to avoid using the Borsuk–Ulam theorem and heavy-duty algebraic-topology arguments if possible. Brouwer’s level is as high as I’d like to preferably reach.)
Any suggestions would be appreciated.
 A: Hopefully another answerer will have an idea about how to do this without a little algebraic topology machinery, but here is an initial solution.
You are correct that $F$ must have a fixed point. Proof by contradiction:
$F$ is defined on the compact set $B^n$, so the image is compact. Thus both $B^n$ and $F(B^n)$ are contained in some sphere $S$ (of dimension $n-1$) of large radius, centered at the origin.
Suppose $F$ has no fixed point. Then (using the same trick in the proof of the Brouwer f.p. theorem) for each $x\in B^n$, there is a unique ray from $x$ to $F(x)$, hitting $S$ in a unique point, and this point varies continuously with $x$. Thus we construct a continuous map $\Phi: B^n\rightarrow S$.
The condition that if $x\in S^{n-1}$ then $F(-x) = -F(x)$ means that the ray from $x$ to $F(x)$ negates the ray from $-x$ to $F(-x)$, and so their intersections with $S$ are antipodes. This means that the restriction of $\Phi$ to the boundary $S^{n-1}$ sends antipodes to antipodes.
Thus if $i:S^{n-1}\hookrightarrow B^n$ is the inclusion, the composed map 
$$S^{n-1}\xrightarrow{i} B^n\xrightarrow{\Phi} S$$
sends antipodes to antipodes.
By a theorem of Borsuk (which is supposed to be equivalent to the Borsuk-Ulam theorem, although I'm afraid I don't know why), a map from the sphere to the sphere that preserves antipodes must have odd degree. In particular, the degree of $\Phi\circ i$ is not zero. So the induced map on homology 
$$H_{n-1}(S^{n-1})\xrightarrow{i_\star} H_{n-1}(B^n) \xrightarrow{\Phi_\star} H_{n-1}(S)$$
is not zero. But this contradicts the fact that $H_{n-1}(B^n) = 0$ (since $B^n$ is contractible).
