Suppose $A$ is an algebra of signature $\mathcal{L}$ and $V=Var(A)$ is the variety generated by $A$. I want to know is it possible to classify relatively free elements of $V$? As a special case, for a group $G$, under what conditions $G$ is free in $Var(G)$?

Suppose that $A$ is a finite universal algebra with minimal cardinality of a generating set $d$. Then $A$ is relatively free in some variety iff it is relatively free on $d$ generators in the variety it generates, in which case it is free on any generating set of $d$ elements. Moreover, this occurs iff each map from a fixed $d$element set generating $A$ to $A$ extends to an endomorphism. It is easy to check that if $A$ is relatively free in some variety then it is relatively free in $Var(A)$. Next note that if $A$ is relatively free on $k$ generators, then the cardinality of $Hom(A,A)$ is $A^k$. On the other hand, $Hom(A,A)\leq A^d$ and so $k\leq d$. But $d$ was minimal, so $A$ is free on $d$ generators. Also note that if $X$ is a free set of $d$generators, then we can map $X$ to any other set of $d$generators and this extends to a surjective homomorphism which is injective by finiteness. So all $d$element generating sets are free generating sets. Clearly if $A$ is relatively free on $d$ generators, then any map from a $d$element generating set (necessarily a free generating set) to $A$ extends to an endomorphism. Suppose the converse holds. Let $B$ be the free algebra on $d$ generators in the variety generated by $A$. It is well known that $B$ embeds in a finite product $A^m$ where $m=A^d$. Let $g_i\colon B\to A$ be the projection to the $i^{th}$ factor. If $f$ is the map taking the $d$ generators of $A$ to the free generators of $B$, then we can extend $g_if$ to an endomorphism $h_i$ of $A$ by hypothesis on $A$. The product $h$ of these $h_i$ gives a homomorphism of $A$ to $A^m$ sending the generators of $A$ to the generators of $B$. Hence $h$ splits the canonical surjection $B\to A$. Added. This last argument doesn't require finiteness. A universal algebra $A$ is relatively free if and ony if it has a generating set $X$ such that each map from $X$ to $A$ extends to an endomorphism. 


More general than my comment above, but still only a partial answer: Notice that $\text{Var}(A)$ is generated by any $B\in\text{Var}(A)$ that has $A$ as a subalgebra (or quotient, or subquotient). It would seem that only in very special varieties would all such $B$'s be free. 


Free algebras of $V(\textbf{A})$ lie in $SP(\textbf{A})$ (subalgebra of product). Unless your algebra is trivial or very special, I do not see a better characterization of free algebras forthcoming. 

