= Solution
{c}
The <Black-Scholes reflection symmetry> becomes a weighted reflection after taking a <logarithm>. Put $x=\log(S/B)$ and $v(x,t)=V(Be^x,t)$. The <multivariable chain rule> transforms the equation into
$$
v_t+\tfrac12\sigma^2v_{xx}+bv_x-rv=0,\qquad b=r-\sigma^2/2.
$$
For $w(x,t)=e^{\gamma x}v(-x,t)$, its left-hand side, after removing the factor $e^{\gamma x}$, is
$$
v_t+\tfrac12\sigma^2v_{xx}-(\sigma^2\gamma+b)v_x
+(\tfrac12\sigma^2\gamma^2+b\gamma-r)v,
$$
evaluated at $(-x,t)$. Taking $\gamma=-2b/\sigma^2$ makes the derivative coefficient $b$ and the additional zeroth-order term zero. Thus
$$
\boxed{W(S,t)=(S/B)^{1-2r/\sigma^2}V(B^2/S,t)}
$$
satisfies the same <Black-Scholes equation>.
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