How does the image of the Hurewicz map $\pi_n(X,x) \to H_n(X)$ depend upon the choice of the base point? Let $X$ be a path connected topological space. I understand that the homotopy groups $\pi_n(X,x_0)$ and $\pi_n(X,x_1)$ are isomorphic to each other. However I do not understand whether the image of the Hurewicz map $\pi_n(X,x) \to H_n(X)$ is dependent or independent of the choice of basepoint. Is there any easy way to understand this ? Apologies if I am asking something sily.
I would greatly appreciate any references. Thanks.

By naturality, it suffices to examine the universal example $X = S^n$.  In this case, the Hurewicz homomorphism $\pi_n(S^n, s) \to H_n(S^n)$ is an isomorphism for any base point $s$, so in particular the image does not depend on the base point.

EDIT: Here is the naturality argument I had in mind.  Let $f \in \pi_n(X, x)$ correspond to a map $f: (S^n, s) \to (X, x)$.  Then we have
$$\require{AMScd}
\begin{CD}
\pi_n(S^n, s) @>\cong>> H_n(S^n) \\
@V{f_*}VV @VV{f_*}V \\
\pi_n(X, x) @>>h> H_n(X)
\end{CD}  
$$
We have $h(f) = f_*(1)$, for $f_*$ the induced map on homology which doesn't care about basepoints.  This reduces the question to the issue of whether the classes $\{f: (S^n, s) \to (X, x_1)\}$ and $\{f: (S^n, s) \to (X, x_2)\}$ induce the same collection of maps in homology.  Homology doesn't care about basepoints, and forgetting base points, we get the same classes of maps up to homotopy provided $x_1$ and $x_2$ are in the same path component of $X$, as $S^n$ is connected for $n > 0$.   
