Deforming antiholomorphic involutions Let $(M,J)$ be a compact smooth almost complex manifold. We can "deform" $J$ as follows: if$A$ is a smooth section of the endomorphism bundle $\mathrm{End}(TM)\to M$ satisfying 
$
AJ=-JA,
$
it follows that 
$
Je^A=e^{-A}J,
$
where $e^A$ is the matrix exponential of $A$. It follows immediately that 
$$
J':=Je^A
$$
is another almost complex structure on $M$. 
Suppose $\phi:M\to M$ is a diffeomorphism which is an anti-$J$-holomorphic involution, i.e. $$
J\circ d\phi=-d\phi \circ J \qquad \text{and} \qquad \phi\circ \phi =\mathrm{id}.
$$
Question.
Can we deform $\phi$ into an anti-$J'$-holomorphic involution? 
I think this will be true if there exist a vector field $\eta\in \Gamma(TM)$ such that the diffeomorphism $\phi':M\to M$ defined by 
$$
\phi'(x)=(\phi\circ \mathrm{exp}\circ \eta)(x)
$$
is an anti-$J'$-holomorphic involution (here $\exp$ is defined with respect to some Riemannian metric on $M$). How can I show the existence/nonexistence of $\eta$?
 A: At least in the 2-dimensional case, your deformations of a given complex structure yield all complex structures on the given oriented surface. (A proof of this is a linear algebra exercise where you analyze endomorphisms of a 2-dimensional real vector space.) However, with few exceptions (among compact surfaces, the only exception is the sphere), a generic complex structure on a surface admits no anti-holomorphic automorphisms (see below). Hence, your $\phi'$ generically does not exist. 
Edit. Let us check that a generic smooth elliptic curve (aka a real flat 2-torus) has no antiholomorphic automorphisms. Let $T^2$ be a flat torus of unit area. An antiholomorphic involution $h$ of $T^2$ is an orientation-reversing isometry of the flat metric. If $h$ has nonempty fixed point set then it is a union of two parallel geodesic circles on $T^2$ which divide $T^2$ in two isometric annuli $A_1, A_2$ (swapped by $h$). The conformal type of $T^2$ is then uniquely determined by the modulus of $A_1$ which is a single real number. Hence, the space of flat tori  of unit area which admit such an involution is real 1-dimensional. The second possibility is that $h$ acts freely on $T^2$ and the quotient is the Klein bottle. The moduli space of flat  Klein bottles of fixed area is again real 1-dimensional. (Proving this is requires a tiny bit of work.) On the other hand, the moduli space of elliptic curves is complex one-dimensional.  Hence, a generic elliptic curve admits no antiholomorphic involutions. (One can actually do better: The moduli space of $T^2$ is the complex line with two marked points corresponding to elliptic curves with groups extra holomorphic symmetries. Then the set of of elliptic curves which admit an antiholomorphic involution is the unique real line passing through these two marked points.) 
The proof in the higher genus case is a similar dimension count. The moduli space of compact Riemann surfaces of genus $g$ has dimension $3g-3$. On the other hand, if $h: S\to S$ is an antiholomorphic involution of a genus $g$ surface with nonempty fixed point set $F$, then $S$ is determined by the conformal  type of one component $C$ of $S-F$. You have that $\chi(S)=2\chi(C)$. Then you compute the dimension of the moduli space of $C$, it equals $3g'-3+2p$ where $g'$ is the genus of $C$ and $p$ is the number of components of $F$. By the Euler characteristic formula above, you get: $2- 2g = 2(2g'-p+2)$. 
Then make a computation and conclude that $3g'-3+2p$ is strictly less than $3g-3$. The proof when $h$ has no fixed points is a similar dimension count. 
