It seems that contrary to some other answers a continuous solution can be constructed.

First of all we interpolate with Newton series the flow of function $\cos(\cos(z))$:

$$\phi_{1/2}(x,z)=\cases {
\arccos^{[x]}(z), & \text{if } x < 0 \cr
\cos^{[x]}(z), & \text{if } x \ge 0 }
$$

$$\phi_{1}(x,z)=\sum_{m=0}^\infty \binom{x/2+1}{m} \sum_{k=0}^m (-1)^{k-m} \binom{m}{k} \phi_{1/2}(k-1,z)$$

We interpolate from the first integer point where the value is real, i.e. from x=-1.

We now obtain the approximation of the other half-flow of cos x by taking arccos on the above function:

$$\phi_{2}(x,z)=\arccos(\phi_{1}(x+1,z))$$

We know that the flow of cos(x) should coincide with the first function in even integers and with the second function in odd integers.

So we make a stub of the flow following this knowledge (we also want that its absolute value to be monotonous).

$$\phi(x,z)=\frac{1}{2} \left((-1)^{x+1}+1\right) (\phi_{1}(x,z)-\text{FP})+\frac{1}{2} \left((-1)^x+1\right) (\phi_{2}(x,z)-\text{FP})+\text{FP}$$

where FP is the cosine fixed point.

This function coincides with the flow in integer points but still disagrees in between.
To get a real flow we have to take a limit of repeated arccosine on the our stub:

$$\Phi(x,z)=\lim_{n\to\infty} \arccos^{[n]} (\phi(x+n,z))$$

Numerically this limit converges quite fast. If the limit exists, it by definition, satisfies the equation

$$ \cos(\Phi(x,z))=\Phi(x+1,z)$$

so it is the true flow.

The above can be illustrated by the graphic:

Here upper semi-flow (flow of cos(cos z)) ) is blue, lower semi-flow is red, real part of the flow is yellow, imaginable part of the flow is green. All flows are taken as point z=1.

Following this we can build a graphic of half-iterate of cosine $\Phi(1/2,z)$:

Here blue is the real part and red is the imaginary part.

We can verify that the half-iterate repeated twice $\Phi(1/2,\Phi(1/2,z))$ (blue) follows cosine (red) quite well at positive half-periods, and anywhere the cone is positive (that is, on the imaginary axis as well):

I think this coincodes with the answer by Gerald Edgar above.
A modified function, iterated twice gives cosine in all real axis:

This is a true half-iterate of cosine, which works on the whole real axis, producing exactly cosine:

But as has been noted by Joel David Hamkins above, there is infinite number of such solutions, none of which work for the whole complex numbers.

This function can be considered though as the true solution on the complex plane if interpreted as a multi-valued function. To do this, take the function on the each interval and analytically extend it to the whole complex plane.

A mathematica notebook that produces the above is as follows:

```
$PlotTheme = None;
f[x_, z_] := If[x >= 0, Nest[Cos, z, 2*x], Nest[ArcCos, z, -2*x]]
n := 30
s := 15
Ni[x_, z_] :=
Sum[Binomial[x + 1, m]*
Sum[(-1)^(k - m)*Binomial[m, k]*f[k - 1, z], {k, 0, m}], {m, 0, n}]
Semi2[x_, z_] := Ni[x/2, z]
Semi1[x_, z_] := ArcCos[Semi2[x + 1, z]]
FP := Evaluate[N[FixedPoint[Cos, 1.]]]
a := 21
Flow2[x_, z_] :=
FP + (Semi2[x, z] - FP)*(((-1)^x + 1)/2) + (Semi1[x, z] -
FP)*(((-1)^(x + 1) + 1)/2)
FL[x_, z_] := Nest[ArcCos, Flow2[x + a, z], a]
Plot[{Semi1[x, 1], Semi2[x, 1], Re[FL[x, 1]], Im[FL[x, 1]]}, {x, -5,
5}, AspectRatio -> Automatic, PlotRange -> 3]
Plot[{Re[FL[0.5, x]], Im[FL[0.5, x]]}, {x, -5, 5},
AspectRatio -> Automatic, PlotRange -> 3]
Plot[{Re[FL[0.5, FL[0.5, x]]], Cos[x]}, {x, -5, 5},
AspectRatio -> Automatic, PlotRange -> 3]
HalfCos[z_] :=
If[Im[z] == 0, Sign[Re[Cos[z]]]*FL[0.5, z], Sign[Re[z]]*FL[0.5, z]]
Plot[{Re[HalfCos[x]], Im[HalfCos[x]]}, {x, -5, 5},
AspectRatio -> Automatic, PlotRange -> 3]
Plot[{Re[HalfCos[HalfCos[x]]], Cos[x]}, {x, -5, 5},
AspectRatio -> Automatic, PlotRange -> 3]
```