Q:

Proving isomorphism between graphs

If I'm asked to prove two graphs are isomorphic by constructing an isomorphism E.g for these two graphs if I start from $u_1$ I have an option to send $u_1$ to any of $v_1$ to $v_6$ and I start by sending $u_1$ to $v_1$ and then for $u_2, u_2$ is adjacent to $u_1$ so image of $u_2$ will be adjacent to image of $u_1$ which is $v_1$ so I have the options of $v_4, v_5$ and $v_6$. 
My question is that won't we then have different isomorphisms because of the different options that we have to send $u_2$ to, also if I had decided to send $u_1$ to $v_2$ at the start instead of to $v_1$ for example then would  have obtained a different isomorphism, so is there not a unique answer for the isomorphism? So basically if two graphs are isomorphic should we be able to construct an isomorphism regardless of which vertex we choose to start from?
I'm asking because the answer that I obtained is different from the one given in the mark scheme,
The answer I obtained for the isomorphism was \begin{align*}
u_1 &\to v_1\\ u_2 &\to v_4\\ u_3 &\to v_2\\ u_4 &\to v_5\\ u_5 &\to v_3\\ u_6 &\to v_6,\end{align*} I would be grateful if someone could check this for me.

A:

Yes, you are correct - you have found an isomorphism.
Since there is some isomorphism between these two graphs, we call them isomorphic, and consider them to be, in some sense, "essentially the same". In general, the collection of isomorphisms between a graph and itself is called the automorphism group of the graph; depending on the graph, there can anywhere from one (trivial) automorphism, to many automorphisms. 
In this case, both graphs are bipartite. In fact, these graphs are both isomorphic to $K_{3, 3}$, the complete bipartite graph whose partitions both have size $3$. There are ${2 \choose 1}\cdot 3!\cdot 3! = 72$ isomorphisms between the two graphs, because of this.
EDIT: To elaborate on one specific question of yours:

So basically if two graphs are isomorphic should we be able to construct an isomorphism regardless of which vertex we choose to start from?

It depends exactly what you mean. In one sense, yes: If two graphs are isomorphic, then you can pick any vertex to start with, but that doesn't mean you can choose to send it to any other vertex. Here, you can, but only because $K_{3, 3}$ is so symmetric. Let's say we start off by choosing to send $u_1 \to v_1$. Then the even-subscripted $u$'s, that is, $u_2, u_4$, and $u_6$, can only be sent to something in $\{v_4, v_5, v_6\}$, by the "bipartite" considerations.

