1. Find the coefficient of $x^7$ in $\left(x^2+\frac{1}{bx}\right)^{11}.$

   The general term is $T_{r+1}=\binom{11}{r}(x^2)^{11-r}\left(\frac{1}{bx}\right)^r.$

   Simplifying,

   =\binom{11}{r}b^{-r}x^{22-3r}.$$ For the power of $x$ to be $7$, $$22-3r=7 \implies 3r=15 \implies r=5.$$ So the coefficient of $x^7$ is $$\binom{11}{5}b^{-5}=\frac{\binom{11}{5}}{b^5}.$$

2. Find the coefficient of $x^{-7}$ in $\left(x-\frac{1}{bx^2}\right)^{11}.$

   The general term is $T_{r+1}=\binom{11}{r}x^{11-r}\left(-\frac{1}{bx^2}\right)^r.$

   Simplifying,

   =\binom{11}{r}(-1)^rb^{-r}x^{11-3r}.$$ For the power of $x$ to be $-7$, $$11-3r=-7 \implies 3r=18 \implies r=6.$$ Hence the coefficient of $x^{-7}$ is $$\binom{11}{6}(-1)^6b^{-6}=\frac{\binom{11}{6}}{b^6}.$$

3. Equate the coefficients

   Given they are equal, $\frac{\binom{11}{5}}{b^5}=\frac{\binom{11}{6}}{b^6}.$

   Since $\binom{11}{5}=\binom{11}{6},$ we get $\frac{1}{b^5}=\frac{1}{b^6}.$

   Multiplying by $b^6$ (given $b\neq 0$), $b=1.$

4. Check options

   • A: $2$ ❌
   • B: $-1$ ❌
   • C: $1$ ✅
   • D: $-2$ ❌

Therefore, the correct answer is C: $1$.