Given to unitary vectors $\bf u$ and $\bf v$, thus two points on the unitary sphere,
there are of course infinite paths leading from $\bf u$ to $\bf v$.
A neat and simple (as much as spherical geometry allows) approach I think is that of taking
the shortest path, i.e. the path along the great-circle or orthodrome.
That will lay on the plane through the origin and parallel to $\bf u$ and $\bf v$.

The unit normal to that plane (Right_Hand rule $\bf u \to \bf v$) is
$$
{\bf n} = {{{\bf u} \times {\bf v}} \over {\left| {{\bf u} \times {\bf v}} \right|}}
$$
It is useful to consider also another unit vector $\bf t$, lying on the plane and normal to $\bf u$
$$
{\bf t} = {\bf n} \times {\bf u}
$$
which already normalized by definition.
Note that the right handed ref. system $x,y,z$ transforms into the right handed ref. system $\bf u, \bf t, \bf n$.
If we put these last three vectors as the columns of a matrix, we obtain
$$
\left( {\begin{array}{c|c|c}
{\bf u} & {\bf t} & {\bf n} \\
\end{array}} \right) = {\bf T}\quad \Rightarrow \quad \left( {\begin{array}{c|c|c}
{\bf u} & {\bf t} & {\bf n} \\
\end{array}} \right) = {\bf T}\left( {\begin{array}{c|c|c}
{\bf i} & {\bf j} & {\bf k} \\
\end{array}} \right)
$$
and $\bf T$ is by definition orthogonal, unitary and with $|\bf T|=1$. Thus it is also $\bf T^{-1}= \bf T^{T}$.
The great circle angle from $\bf u$ to $\bf v$ is given, in absolute value and sign by
$$
\alpha = \arctan _2 \left( {{\bf v} \cdot {\bf u},\;{\bf v} \cdot {\bf t}} \right) = \arctan _2 \left( {\cos \alpha ,\;\sin \alpha } \right)
$$
that is we will use both the cosine and sine value to determine $\alpha$.
So $(\cos \alpha, \sin \alpha,0)$ are the coordinates of $\bf v$ in the $(\bf u, \bf t, \bf n)$ sytem, that is we can put
$$
{\bf v} = {\bf R}_{\,{\bf n}} (\alpha )\;{\bf u} = \left( {\matrix{
{\cos \alpha } & { - \sin \alpha } & 0 \cr
{\sin \alpha } & {\cos \alpha } & 0 \cr
0 & 0 & 1 \cr
} } \right)\;{\bf u}
$$
Since we can write
$$
\begin{array}{l}
{\bf v} = \left( {\begin{array}{*{20}c}
{\bf u} &| & {\bf t} &| & {\bf n} \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
{v_{\,u} } \\ {v_{\,t} } \\ {v_{\,n} } \\
\end{array}} \right) = {\bf T}\left( {\begin{array}{*{20}c}
{\bf i} &| & {\bf j} &| & {\bf k} \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
{v_{\,u} } \\ {v_{\,t} } \\ {v_{\,n} } \\
\end{array}} \right) = \\
= {\bf T}\;\left( {\begin{array}{*{20}c}
{\bf i} &| & {\bf j} &| & {\bf k} \\
\end{array}} \right)\;{\bf T}^{\, - \,{\bf 1}} \;{\bf T}\;\left( {\begin{array}{*{20}c}
{v_{\,u} } \\ {v_{\,t} } \\ {v_{\,n} } \\
\end{array}} \right) = {\bf T}\;{\bf I}\;{\bf T}^{\, - \,{\bf 1}} \;{\bf T}\;\left( {\begin{array}{*{20}c}
{v_{\,u} } \\ {v_{\,t} } \\ {v_{\,n} } \\
\end{array}} \right) = \\
= \;{\bf T}\;\left( {\begin{array}{*{20}c}
{v_{\,u} } \\ {v_{\,t} } \\ {v_{\,n} } \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
{\bf i} &| & {\bf j} &| & {\bf k} \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
{v_{\,x} } \\ {v_{\,y} } \\ {v_{\,z} } \\
\end{array}} \right) \\
\end{array}
$$
then
$$
\begin{array}{l}
\left( {\begin{array}{*{20}c}
{v_{\,x} } \\
{v_{\,y} } \\
{v_{\,z} } \\
\end{array}} \right) = \;{\bf T}\;\left( {\begin{array}{*{20}c}
{v_{\,u} } \\
{v_{\,t} } \\
{v_{\,n} } \\
\end{array}} \right) = {\bf TR}_{\,{\bf n}} (\alpha )\;{\bf u}_{u,\,t,\,n} = {\bf TR}_{\,{\bf n}} (\alpha )\;{\bf T}^{\, - \,{\bf 1}} \;{\bf T}\;{\bf u}_{u,\,t,\,n} = \\
= {\bf TR}_{\,{\bf n}} (\alpha )\;{\bf T}^{\, - \,{\bf 1}} \;{\bf T}\;{\bf u}_{x,\,y,\,z} \\
\end{array}
$$
and finally, with $\bf u, \; \bf v$ expressed in the $(x,y,z)$ ref. system:
$$
{\bf v} = {\bf T}\,{\bf R}_{\,{\bf n}} (\alpha )\;{\bf T}^{\, - \,{\bf 1}} \,{\bf u}
$$
As you can see it is a quite fluid and general procedure.
As for the example you give, we have
$$
\begin{array}{l}
{\bf u} = \left( {\begin{array}{*{20}c}
1 \\ 0 \\ 0 \\
\end{array}} \right)\quad \quad {\bf v} = \left( {\begin{array}{*{20}c}
0 \\ {\sqrt 2 /2} \\ {\sqrt 2 /2} \\
\end{array}} \right) \\
{\bf n} = \frac{{{\bf u} \times {\bf v}}}{{\left| {{\bf u} \times {\bf v}} \right|}} = \left( {\begin{array}{*{20}c}
0 \\ { - \sqrt 2 /2} \\ {\sqrt 2 /2} \\
\end{array}} \right)\quad \quad {\bf t} = {\bf n} \times {\bf u} = \left( {\begin{array}{*{20}c}
0 \\ {\sqrt 2 /2} \\ {\sqrt 2 /2} \\
\end{array}} \right) \\
\alpha = \arctan _2 \left( {{\bf v} \cdot {\bf u},\;{\bf v} \cdot {\bf t}} \right) = \arctan _2 \left( {0,\;1} \right) = \pi /2 \\
{\bf T} = \left( {\begin{array}{*{20}c}
{\bf u} &| & {\bf t} &| & {\bf n} \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
1 & 0 & 0 \\ 0 & {\sqrt 2 /2} & { - \sqrt 2 /2} \\ 0 & {\sqrt 2 /2} & {\sqrt 2 /2} \\
\end{array}} \right) \\
{\bf R}_{\,{\bf n}} (\alpha ) = \left( {\begin{array}{*{20}c}
{\cos \alpha } & { - \sin \alpha } & 0 \\ {\sin \alpha } & {\cos \alpha } & 0 \\ 0 & 0 & 1 \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
0 & { - 1} & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\
\end{array}} \right) \\
{\bf T}\,{\bf R}_{\,{\bf n}} (\alpha )\;{\bf T}^{\, - \,{\bf 1}} = \\
= \left( {\begin{array}{*{20}c}
1 & 0 & 0 \\ 0 & {\sqrt 2 /2} & { - \sqrt 2 /2} \\ 0 & {\sqrt 2 /2} & {\sqrt 2 /2} \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
0 & { - 1} & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
1 & 0 & 0 \\ 0 & {\sqrt 2 /2} & {\sqrt 2 /2} \\ 0 & { - \sqrt 2 /2} & {\sqrt 2 /2} \\
\end{array}} \right) = \\
= \left( {\begin{array}{*{20}c}
0 & { - \sqrt 2 /2} & { - \sqrt 2 /2} \\ {\sqrt 2 /2} & {1/2} & { - 1/2} \\ {\sqrt 2 /2} & { - 1/2} & {1/2} \\
\end{array}} \right) \\
\end{array}
$$
and in fact
$$
\left( {\begin{array}{*{20}c}
0 \\ {\sqrt 2 /2} \\ {\sqrt 2 /2} \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
0 & { - \sqrt 2 /2} & { - \sqrt 2 /2} \\ {\sqrt 2 /2} & {1/2} & { - 1/2} \\ {\sqrt 2 /2} & { - 1/2} & {1/2} \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
1 \\ 0 \\ 0 \\
\end{array}} \right)
$$
Going your way instead, just for the cited example, you shall first rotate around $z$ by $+\pi/2$ (RH rule),
and then around the new $y$ axis, which has become the $-x$, i.e. $x$ axis, by $+\pi/4$ (RH rule again).
The matrices are
$$
{\bf R}_{\,{\bf x}} (\alpha ) = \left( {\begin{array}{*{20}c}
1 & 0 & 0 \\
0 & {\cos \alpha } & { - \sin \alpha } \\
0 & {\sin \alpha } & {\cos \alpha } \\
\end{array}} \right)\quad {\bf R}_{\,{\bf y}} (\beta ) = \left( {\begin{array}{*{20}c}
{\cos \beta } & 0 & {\sin \beta } \\
0 & 1 & 0 \\
{ - \sin \beta } & 0 & {\cos \beta } \\
\end{array}} \right)\quad {\bf R}_{\,{\bf z}} (\gamma ) = \left( {\begin{array}{*{20}c}
{\cos \gamma } & { - \sin \gamma } & 0 \\
{\sin \gamma } & {\cos \gamma } & 0 \\
0 & 0 & 1 \\
\end{array}} \right)
$$
they always act on the base reference, and the sign of the angle is according to the RH rule, so
$$
\begin{array}{l}
\left( {\begin{array}{*{20}c}
0 \\ {\sqrt 2 /2} \\ {\sqrt 2 /2} \\
\end{array}} \right) = {\bf R}_{\,{\bf x}} (\pi /4){\bf R}_{\,{\bf z}} (\pi /2)\left( {\begin{array}{*{20}c}
1 \\ 0 \\ 0 \\
\end{array}} \right) = \\
= \left( {\begin{array}{*{20}c}
1 & 0 & 0 \\ 0 & {\sqrt 2 /2} & { - \sqrt 2 /2} \\ 0 & {\sqrt 2 /2} & {\sqrt 2 /2} \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
0 & { - 1} & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
1 \\ 0 \\ 0 \\
\end{array}} \right) = \\
= \left( {\begin{array}{*{20}c}
0 & { - 1} & 0 \\ {\sqrt 2 /2} & 0 & { - \sqrt 2 /2} \\ {\sqrt 2 /2} & 0 & {\sqrt 2 /2} \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
1 \\ 0 \\ 0 \\
\end{array}} \right) \\
\end{array}
$$