1. Given relation

We have $B=\begin{bmatrix}1&3\\1&5\end{bmatrix}$ and $AB^{-1}=A^{-1}.$

We want to use this to find a polynomial satisfied by $A$, then by $C$.

2. Derive an equation for $A$

From $AB^{-1}=A^{-1},$ multiply on the right by $B$: $A=A^{-1}B.$ Now multiply on the left by $A$: $A^2=B.$

So $A$ is a square root of $B$.

3. Use similarity of $A$ and $C$

Given $BCB^{-1}=A,$ this means $A$ and $C$ are similar matrices. Hence they satisfy the same characteristic polynomial, and any polynomial equation satisfied by $A$ is also satisfied by $C$.

Since $A^2=B$, let us first find a polynomial satisfied by $B$.

4. Find the characteristic polynomial of $B$

For $B=\begin{bmatrix}1&3\\1&5\end{bmatrix},$ its trace and determinant are $\operatorname{tr}(B)=1+5=6,$ $\det(B)=1\cdot 5-3\cdot 1=2.$

Therefore the characteristic polynomial is $\lambda^2-6\lambda+2.$

By Cayley-Hamilton, $B^2-6B+2I=O.$

5. Convert this into an equation for $A$

Since $B=A^2$, substitute into the above: $(A^2)^2-6A^2+2I=O,$ that is, $A^4-6A^2+2I=O.$

Because $C$ is similar to $A$, $C$ satisfies the same polynomial: $C^4-6C^2+2I=O.$

Compare with $C^4+\alpha C^2+\beta I=O.$ So, $\alpha=-6, \qquad \beta=2.$

6. Compute $2\beta-\alpha$

$2\beta-\alpha=2(2)-(-6)=4+6=10.$

7. Check options

The correct option is: $\boxed{10}$ which is Option B.

8. Comparison with stored answer

Stored correct answer: B

Our derived answer is also B. So they agree.