The finite and infinite cases are actually quite different for this problem. In the simpler version of the finite case where $X=Y$ and $\mu=\nu$, you can show that the extreme points are introduced by maps $f: X\to X$ such that $\mu(f(x)) = \mu(x)$ almost everywhere, so for each distinct positive mass assumed by an atom under $\mu$, $f$ permutes the atoms with this mass.
In the infinite case, not all extreme points can be induced in this way. For an example, take $X = Y = [0,1/2]$ and let $M$ be the sum of one-dimensional Lebesgue measure on the lines $x = 1/4$ and $y=1/4$. Then $M$ has equal marginals, namely Lebesgue measure $\lambda$ plus a point mass $\delta$ of weight $1/2$ at $x = 1/4$.
To see that $M$ is extreme, write $M = pM_1 + (1-p)M_2$ for $M_1,M_2\in C(\lambda + \delta,\lambda + \delta)$ and $p\in [0,1]$. By the Radon-Nikodym theorem, the $M_i = u_iM$ for some density $u_i$. Since $M$ is supported on the two line segments, $M$-almost everywhere $u_i(x,y) = v_i(x) + w_i(y)$ for some $v_i$ and $w_i$ with $v_i(1/2) = w_i(1/2) = 0$. The marginals of $u_i M$ are $v_i\lambda + \left[2\int w_i(y)dy\right]\delta$ and $w_i\lambda + \left[2\int v_i(y)dy\right]\delta$. Since $M_1,M_2\in C$, these must be equal to $\lambda + \delta$, so $v_i = w_i = 1$ $\lambda$-almost everywhere. That is, $M_1 = M_2 = M$ and $M$ is extreme.
