Let $\mathfrak{M}$ be a model, $A \subset M$ and $a,b \in M$. Assume $\text{tp}(a/A) = \text{tp}(b/A)$. Does this imply $\text{tp}(a/\text{acl}(A)) = \text{tp}(b/\text{acl}(A))$?
I feel that this statement should be true, and I have tried to prove it as follows: Let $\varphi(x,c_1,\ldots,c_n) \in \text{tp}(a/\text{acl}(A))$ where $c_1,\ldots,c_n$ are the parameters in $\text{acl}(A)$. Choose for every $c_i$ an algebraic formula $\psi_{c_i} \in L(A)$ such that $c_i \in \psi_{c_i}(\mathfrak{M})$ and $\psi_{c_i}(\mathfrak{M})$ has minimal cardinality. I would like to show that
(*) $\varphi(x,c_1,\ldots,c_n) \leftrightarrow \exists y_1 \ldots \exists y_n \varphi(x,y_1,\ldots,y_n)\wedge \psi_{c_1}(y_1) \wedge \ldots \wedge \psi_{c_1}(y_n)$
in $(\mathfrak{M},\text{acl}(A))$. Then I could use my hypothesis. However I don't know how to derive the statement (*) (or how to find a counter-example).
The statement is false unless $acl(A) = dcl(A)$. If $a \in acl(A) \setminus dcl(A)$, then you can find $b \neq a$ in $acl(A)$ that realises the same type over $A$. (Hint: look at realisations of the formula $\phi(x) \in tp(a/A)$ that defines a set of least cardinality.) Now since $a, b \in acl(A)$ clearly $tp(a/acl(A)) \neq tp(b/acl(A))$.
On the other hand $tp(a/A) = tp(b/A)$ implies $tp(a/dcl(A)) = tp(b/dcl(A))$. This follows from the fact that every set definable over $dcl(A)$ is definable over $A$ already.