[proofplan] One direction is immediate from elementarity: an automorphism fixing $A$ preserves the truth of every formula with parameters from $A$. For the converse, equality of complete types over $A$ makes the coordinate map $a_i \mapsto b_i$ compatible with all formulas over $A$, including equality relations and possible coincidences with elements of $A$. Hence this coordinate map together with the identity on $A$ is a small partial elementary map, and the homogeneity of the monster model extends it to an automorphism of $\mathfrak C$ fixing $A$ pointwise. [/proofplan]

[step:Use an automorphism fixing $A$ to preserve all formulas over $A$] Assume that $\sigma \in \operatorname{Aut}(\mathfrak C/A)$ satisfies $\sigma(a)=b$. Let $\varphi(x_1,\dots,x_n,c_1,\dots,c_m)$ be an $L$-formula, where $m \in \mathbb N$ and $c_1,\dots,c_m \in A$. Since $\sigma$ fixes each $c_j$ and is elementary, \begin{align*} \mathfrak C \models \varphi(a_1,\dots,a_n,c_1,\dots,c_m) \iff \mathfrak C \models \varphi(\sigma(a_1),\dots,\sigma(a_n),\sigma(c_1),\dots,\sigma(c_m)). \end{align*} Using $\sigma(a)=b$ and $\sigma(c_j)=c_j$ for each $1 \leq j \leq m$, this becomes \begin{align*} \mathfrak C \models \varphi(a_1,\dots,a_n,c_1,\dots,c_m) \iff \mathfrak C \models \varphi(b_1,\dots,b_n,c_1,\dots,c_m). \end{align*} Thus $a$ and $b$ satisfy exactly the same formulas with parameters from $A$, so \begin{align*} \operatorname{tp}(a/A)=\operatorname{tp}(b/A). \end{align*} [/step]

[step:Build the partial elementary map from equality of types]Assume \begin{align*} \operatorname{tp}(a/A)=\operatorname{tp}(b/A). \end{align*} Define the small subset \begin{align*} D := A \cup \{a_1,\dots,a_n\} \subset \mathfrak C. \end{align*} We define a map $f:D \to \mathfrak C$ by \begin{align*} f(c) &:= c &&\text{for } c \in A,\\ f(a_i) &:= b_i &&\text{for } 1 \leq i \leq n. \end{align*} This definition is well-defined. If $a_i=a_j$, then the formula $x_i=x_j$ belongs to $\operatorname{tp}(a/A)$, hence belongs to $\operatorname{tp}(b/A)$, so $b_i=b_j$. If $a_i=c$ for some $c \in A$, then the formula $x_i=c$ belongs to $\operatorname{tp}(a/A)$, hence belongs to $\operatorname{tp}(b/A)$, so $b_i=c$. We claim that $f$ is partial elementary. Let $r \in \mathbb N$, let $d_1,\dots,d_r \in D$, and let $\psi(y_1,\dots,y_r)$ be an $L$-formula. For each $1 \leq k \leq r$, either $d_k \in A$ or $d_k=a_i$ for some $1 \leq i \leq n$. Replacing the entries from $A$ by parameters and the entries among $a_1,\dots,a_n$ by their corresponding variables, the truth of $\psi(d_1,\dots,d_r)$ is equivalent to the truth of an $L(A)$-formula in the tuple $a$. Since $a$ and $b$ have the same complete type over $A$, the same formula holds of $b$. Translating back gives \begin{align*} \mathfrak C \models \psi(d_1,\dots,d_r) \iff \mathfrak C \models \psi(f(d_1),\dots,f(d_r)). \end{align*} Therefore $f:D \to f(D)$ is a partial elementary map.[/step]

[guided]We need to turn syntactic equality of types into an actual map. The intended map fixes the parameter set $A$ and sends each coordinate $a_i$ to the corresponding coordinate $b_i$. Define \begin{align*} D := A \cup \{a_1,\dots,a_n\}. \end{align*} Then set $f(c):=c$ for $c \in A$ and $f(a_i):=b_i$ for $1 \leq i \leq n$. There are two possible well-definedness issues. First, two coordinates of $a$ might be equal. If $a_i=a_j$, then $\mathfrak C \models x_i=x_j$ at $a$, so the formula $x_i=x_j$ lies in $\operatorname{tp}(a/A)$. Since the two types over $A$ are equal, this formula also lies in $\operatorname{tp}(b/A)$, and therefore $b_i=b_j$. Second, a coordinate of $a$ might already be a parameter from $A$. If $a_i=c$ for some $c \in A$, then the $L(A)$-formula $x_i=c$ lies in $\operatorname{tp}(a/A)$, hence lies in $\operatorname{tp}(b/A)$, so $b_i=c$. Thus the two defining rules for $f$ never conflict. Now we verify that $f$ is partial elementary. Let $r \in \mathbb N$, let $d_1,\dots,d_r \in D$, and let $\psi(y_1,\dots,y_r)$ be any $L$-formula. Each $d_k$ is either a parameter from $A$ or one of the coordinates $a_i$. By replacing the parameters from $A$ by named parameters and replacing each occurrence of $a_i$ by the variable $x_i$, the assertion \begin{align*} \mathfrak C \models \psi(d_1,\dots,d_r) \end{align*} becomes the assertion that some $L(A)$-formula $\theta(x_1,\dots,x_n)$ holds of $a$. Since $\operatorname{tp}(a/A)=\operatorname{tp}(b/A)$, we have \begin{align*} \mathfrak C \models \theta(a_1,\dots,a_n) \iff \mathfrak C \models \theta(b_1,\dots,b_n). \end{align*} Translating this statement back through the definition of $f$ gives \begin{align*} \mathfrak C \models \psi(d_1,\dots,d_r) \iff \mathfrak C \models \psi(f(d_1),\dots,f(d_r)). \end{align*} Since this holds for every finite tuple from $D$ and every $L$-formula, the map $f:D \to f(D)$ is partial elementary.[/guided]

[step:Extend the partial elementary map by homogeneity] The set $D=A\cup\{a_1,\dots,a_n\}$ is small because $A$ is small and $n$ is finite. Since $\mathfrak C$ is sufficiently homogeneous, every partial elementary map with small domain extends to an automorphism of $\mathfrak C$. Applying this to $f:D \to f(D)$, there exists $\sigma \in \operatorname{Aut}(\mathfrak C)$ such that $\sigma|_D=f$. Because $f(c)=c$ for every $c \in A$, the extension $\sigma$ fixes $A$ pointwise, so \begin{align*} \sigma \in \operatorname{Aut}(\mathfrak C/A). \end{align*} Because $f(a_i)=b_i$ for each $1 \leq i \leq n$, we also have \begin{align*} \sigma(a)=(\sigma(a_1),\dots,\sigma(a_n))=(b_1,\dots,b_n)=b. \end{align*} Thus there exists $\sigma \in \operatorname{Aut}(\mathfrak C/A)$ with $\sigma(a)=b$. [/step]

[step:Conclude the equivalence] The first step proves that every automorphism in $\operatorname{Aut}(\mathfrak C/A)$ sending $a$ to $b$ forces \begin{align*} \operatorname{tp}(a/A)=\operatorname{tp}(b/A). \end{align*} The second and third steps prove that equality of these complete types produces such an automorphism. Therefore \begin{align*} \operatorname{tp}(a/A)=\operatorname{tp}(b/A) \iff \exists \sigma \in \operatorname{Aut}(\mathfrak C/A)\ \sigma(a)=b. \end{align*} This is the desired equivalence. [/step]