Multiplicity of a dual simple module in the dual module? Let $A$ be a finite dimensional $k$ algebra. 
Let $S$ be a simple left $A$-module and $M$ be any left $A$ module.
Then my first question is that is it true
$$[M:S]=[M^*:S^*]$$ where $(-)^*=Hom_k(-,k)$  
Second question, under what condition $_AS^*\simeq _AS $
I would be thankful for any reference or hint for these questions. 
**the composition
multiplicity $[M : S]$ counting the number of factors of a composition
series of $M$ that are isomorphic to S.
This is my question asked as guest, that is why I edited it. 
 A: First, for a left $A$-module $M$ the dual $M^*$ is naturally a right $A$-module, or equivalently, a module over the opposite algebra $A^\mathrm{op}$. If $A$ is equipped with an anti-automorphism you can use it to turn $M^*$ into a left $A$-module again (this happens for instance with group algebras and more generally with Hopf algebras). 
I'll give two proofs of the first fact you asked about.
If $0=M_0 \subseteq M_1 \subseteq \cdots \subseteq M_\ell=M$ is a composition series for $M$, we define submodule $M_i^\perp$ of $M^*$ by
$$M_i^\perp=\{f \in M^* \ | \ f(M_i)=0 \}.$$ Thus we have a sequence $0=M_\ell^\perp \subseteq M_{\ell-1}^\perp \subseteq \cdots \subseteq M_0^\perp=M^*$ of submodules of $M$. Restriction defines an isomorphism $M^*/M_i^\perp \cong M_i^*$ and $(M_i/M_{i-1})^*$ is identified with the submodule of $M_i^*$ consisting of functions vanishing on $M_{i-1}$. Thus $(M_i/M_{i-1})^* \cong M_{i-1}^\perp / M_i^\perp$. Taking into account the fact that the dual of a simple module is simple this proves $[M:S]=[M^*:S^*]$.
Now a quick conceptual proof of the equality $[M:S]=[M^*,S^*]$ goes as follows (assuming we are working over an algebraically closed field $F$): the left hand side may be computed as the dimension $$[M:S]=\mathrm{dim}_F(\mathrm{Hom}_A(P(S),M))$$ where $P(S)$ is the projective cover of $S$ while the right hand side is
$$[M^*:S^*]=\mathrm{dim}_F(\mathrm{Hom}_{A^\mathrm{op}}(M^*,I(S^*)))$$ where $I(S^*)$ is the injective hull of $S^*$. Since $P(S)^*=I(S^*)$ this does it. 
I don't know a simple answer to your second question (that is, assuming $A$ is equipped with an anti-automorphism so that $S^*$ becomes a left $A$-module). 
