Let $A$ be a finite dimensional algebra and $M$ a right $A$-module. Then $M$ is a left $B=End\_A(M)$-module.
$M$ is said to have the double centralizer property in case the canonical map given by right multiplication $F: A \rightarrow End\_{B^{op}}(M)^{op}$ is an isomorphism (of algebras). That $F$ is injective means that $M$ is faithful as an $A$-module, meaning $Ma \neq 0$ for all nonzero $a \in A$.
The most famous situation is when $M=fA$ for some idempotent $f$, such that $fA$ is minimal faithful projective-injective. Examples include Schur algebras and blocks of category $\mathcal{O}$.
Questions: 
Let $M=fA$ for some idempotent $f$ (and add some properties if needed, such as M being injective or...)


* Is there an example where the two algebras $A$ and $End\_{B^{op}}(M)^{op}$ are isomorphic as algebras but there is no double centralizer property?
(answer is no,see comments in the answer)
* Is there an example where the two algebras $A$ and $End\_{B^{op}}(M)^{op}$ have the same dimension as a vector-space but there is no double centralizer property? (Answer is yes: see the answer by Dag Oskar Madsen.)



Double centralizer properties
An example for the second question: Let $A$ be the $4$-dimensional algebra $$A=\begin{pmatrix} \Bbbk & 0\\ \Bbbk & \Bbbk \end{pmatrix} \times \Bbbk$$ and $$f=(\begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix},0)$$ Then $B={\rm{End}}\_A(fA) \cong \Bbbk$ and $${\rm{End}}\_{B^{\rm{op}}}(fA)^{\rm{op}}={\rm{End}}\_{\Bbbk}(\Bbbk^2) \cong M\_2(\Bbbk),$$ which is also $4$-dimensional but not isomorphic to $A$.


