= Solution
Assume the invariant set $C$ is nonempty, as required for any fixed-point assertion. For $x\in C$, put $r(x)=\sup_{c\in C}\|x-c\|$ and $r_*=\inf_{x\in C}r(x)$. Boundedness makes these finite. The parallelogram identity gives, for $x,y\in C$,
$$
r\left(\frac{x+y}{2}\right)^2
\leq\frac{r(x)^2+r(y)^2}{2}-\frac{\|x-y\|^2}{4}.
$$
The midpoint lies in $C$, so a minimizing sequence $x_n$ satisfies
$$
\|x_n-x_m\|^2\leq2r(x_n)^2+2r(x_m)^2-4r_*^2\longrightarrow0.
$$
Since $C$ is closed in a <Hilbert space>, the sequence converges to $x_*\in C$. The function $r$ is 1-Lipschitz, hence $r(x_*)=r_*$. The same midpoint inequality proves uniqueness of this minimizing center.
An affine isometry from the <group> maps $C$ onto itself, since its inverse also preserves $C$. Therefore it preserves $r$ and sends its unique minimizer to another minimizer. Uniqueness gives
$$
\boxed{gx_*=x_*\quad\text{for every group element }g.}
$$
This proof applies to a closed bounded convex subset, the substantive meaning of the printed “convex subspace”; if it is literally a bounded <vector> subspace, it is just the zero subspace.
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