The result is perfectly well defined: the projection is the zero vector. The direction of the result is undefined, because the zero vector doesn't have a direction. This is completely analogous to the following question:
"When I subtract $x$ from $a$, is the result defined when $x = a$?"
The answer is "Sure. It's zero. But the answer doesn't have a sign, because only positive and negative numbers have a sign, but zero does not."
As for the projection of one vector on another when the angle is 45 degrees, the answer is "no". The projection of $a$ on $b$ points in the direction of $b$, but the projection of $b$ on $a$ points in the direction of $a$. You might have meant to ask "Are their lengths the same"? The answer there is "no" as well.
Consider a long vector pointing east and a short vector pointing north-east. The projection of the first on the second will be quite long, while the projection of the second on the first will be short (at least using the definition of projection that I like. You could check with your own definition: let $a = (10, 0)$ and $b = (1, 1)$.
IF the vectors have the same length and are 45 degrees apart, then the lengths of the two projections will be equal. But in fact, if the vectors have the same lengths, the lengths of their projections on each other will be equal, regardless of the angle.
(I know this an old post, but I hope to provide clarification for any future visitors.)
If I understand the question correctly, I think it would be clearer to write the vector as
\begin{equation}
\vec{A}=A_r\hat r + A_\theta \hat \theta + A_z \hat z.
\end{equation}
A diagram of the unit vectors $\hat{r}$, $\hat{\theta}$, $\hat{z}$ in cylindrical coordinates can be found here.
If the accepted answer (stating that $|\vec{A}|=\sqrt{A_r^2+A_z^2}$ ) were correct, then any vector that points in the $\hat{\theta}$ direction (including $\hat{\theta}$ itself!) would have a norm of $0$, which is absurd.
The key concept here is that the components $A_r$, $A_\theta$, and $A_z$ tell you how much of $\vec{A}$ points in the direction of each unit vector. Since $\hat{r}$, $\hat{\theta}$, and $\hat{z}$ are orthonormal, $|\vec{A}|=\sqrt{A_r^2+A_\theta^2+A_z^2}$ (as the OP correctly guessed).
If you don't believe me, we can derive this result by expressing the cylindrical unit vectors in terms of the Cartesian unit vectors (see the link from above for details):
\begin{align}
\hat{r}&=\cos\theta\,\hat{x}+\sin\theta\,\hat{y}\\
\hat{\theta}&=-\sin\theta\,\hat{x}+\cos\theta\,\hat{y}\\
\hat{z}&=\hat{z}
\end{align}
This allows us to evaluate the norm in the familiar Cartesian coordinate system:
\begin{align}
\left|\vec{A}\right|^2 &= \left| A_u(\cos\theta\,\hat{x}+\sin\theta\,\hat{y})
+A_\theta(-\sin\theta\,\hat{x}+\cos\theta\,\hat{y}) + A_z\hat{z}\right|^2\\
&= \left| (A_u\cos\theta-A_\theta\sin\theta\,)\hat{x}
+(A_u\sin\theta+A_\theta\cos\theta)\hat{y} + A_z\hat{z}\right|^2\\
&= (A_u\cos\theta-A_\theta\sin\theta)^2 + (A_u\sin\theta+A_\theta\cos\theta)^2 + A_z^2\\
&= A_u(\sin^2\theta+\cos^2\theta) + A_\theta(\sin^2\theta+\cos^2\theta) + A_z^2\\
&=A_u^2+A_\theta^2+A_z^2.
\end{align}
Be careful not to confuse all of this with the common practice of using $(r,\theta,z)$ to represent the location of a point in cylindrical coordinates, in which case the distance from the origin is $\sqrt{r^2+z^2}$. But this is different than the question that was asked, since this $\theta$ represents an angle with respect to the positive $x$-axis, not the component of a vector in the $\hat{\theta}$ direction.
Best Answer
In polar coordinates $(\rho,\phi)$ we write the radial vector
$$\vec r=\hat xx+\hat yy=\hat \rho \rho$$
In spherical coordinates $(r,\theta,\phi)$, we have
$$\vec r=\hat rr$$
The directions are embedded in the unit vectors $\hat \rho=\hat x\cos(\phi)+\hat y\sin(\phi)$ and $\hat r=\hat x\sin(\theta)\cos(\phi)+\hat y\sin(\theta)\sin(\phi)+\hat z\cos(\theta)$. So, if we wish to be more explicit, we can write in polar coordinates
$$\vec r=\hat \rho(\phi)\rho \tag 1$$
and in spherical coordinates
$$\vec r =\hat r(\theta,\phi)r \tag 2$$
where in $(1)$, we see the functional dependence on $\phi$ appear in the unit vector $\hat \rho$ and in $(2)$ we see the functional dependence on $\theta$ and $\phi$ appear in the unit vector $ \hat r$.