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$A$ is said to be similar to $B$ when there exists an invertible matrix $P$ such that $B = P^{-1}AP$.

My question is, is $P$ unique?
Will there only ever be one possible $P$ that makes this statement true?

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    No. Let be $A = B$. Then each matrix $P = \text{diag}(\lambda,\dots,\lambda)$ with positive $\lambda$ suits to your definition. – 2017-02-28
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    Another sort of example: $A$ is similar to itself. $A = P^{-1} A P $ simply means $A$ commutes with $P$. – 2017-02-28
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    How about $A=B=0$, then every invertible $P$ works. – 2017-02-28

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If you find an invertible matrix $C$ which commutes with $A$, i.e. $AC=CA$, then instead of $P$ you can also take $P'=CP$:

\begin{align} P'^{-1}AP' &= (CP)^{-1}A(CP) \\&=P^{-1}C^{-1}ACP\\ &=P^{-1}C^{-1}CAP\\ &=P^{-1}AP=B. \end{align}

As multiples $\lambda I$ of the identity matrix $I$ always commute with $A$, all $\lambda IP=\lambda P$ for any real value $\lambda\not=0$ will do it too.

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$A=I$ is similar to $B=I$; take any invertible matrix $P$ and $B=P^{-1}AP$.